Farhad Rouhbakhsh

Good night

@BalarkaSen @TedShifrin Thanks for your guidance guys. I learned a lot from you. I wish all of my professors or teacher assistants or even my friends here in Iran were like you too, helping and open to silly questions XD.

Nice description

@robjohn what is that?

But it seems I got my answer, nevertheless. The "character" of the examples are like this, having infinite cusps, regardless of how you write their smooth parametrization. Perhaps it is "easier" to write for the zig zag. But I don't know how to parametrize such that the zig zag stays smooth when changing direction. The reason I chose semi-circles instead of zigzag was that I thought with myself that circles "feel" more smooth.

hahaha perfect

yes, thanks for giving me the direction. The point is, I don't know how on the world to parametrize a curve like this. I don't know even how to start. Yeah, analytically it's hard.

@BalarkaSen Pardon for asking, but what do you mean by "modelling each cusp"? Do you mean writing an anylytical definition for it in each interval [1/n, 1/(n+1)], as for example we write "analytically" $(t^3, t^2)$ for for example t in [-1,1] and in the image of the curve we get a cusp at t = 0?

@BalarkaSen I want to have a smooth parametrization of this curve (the image I sent) such that for all natural number n, the derivative at a point t in [0,1] such that γ(t) =(1/n,0) is equal to (0,0). Is it doable?

@TedShifrin Yes professor, but I feel I didn't get it fully. I need to reread your answers and contemplate on them. Fault on my side

@Balarka Obviously, if the tangent line exists at every point, like the case of (cos(t^2), sin(t^2)), the answer is that WE CAN ALWAYS DO THE PARTITION. How about if it doesn't exist in one point?

@BalarkaSen Ok, you are talking about the existence of the tangent line. What I ask is about the result of the problem. That is, is there an example in which $γ(t)$ is smooth, derivatives match up, but γ'(p) = (0,0) AND that we CAN NOT partition it to finite arcs, each of which are smooth functions of one variable?

Define $γ(t)$ as you did in your comment.

@BalarkaSen So far so good. Now the point where I get deeply confused is this: suppose we are given a curve $γ(t)$, but this time with a non-regular parametrization. That is, the curve is smooth with this parametrization, and derivatives of every order match up at 0 and 1, but there exists a point like p in [0,1] such that γ'(p) = (0,0). Then what happens? I want to know your comments up to here and after that I ask my other questions in this regard.

@BalarkaSen Perfect. I understood. So, if these two statements are equivalent, then I think in the problem we have assumed the \textbf{first part of} the first statement ,i.e, that the parametrized curve is infinitely differentiable and the higher order derivatives match up at 0 and 1. And then, adding the \textbf{second part}, that is, the regularity condition, we can use statement 2, that says at each point there exists a tangent line. And then of course the solution of the professor works.

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@BalarkaSen OK I read this. Thanks Balarka for clarification. In the second part, when you say "such that U∩C is a graph of a smooth function defined on ℓ∩U" what you mean is considering ℓ as diffeomorphic to an interval on the y-axis or x-axis

@TedShifrin Ok professor.

@shintuku Bro, read my message. Why did you at the first place intervene in a thing which was not related to you? It was an issue between me and Ted, and finished peacefully. Don't continue.

And I never insulted you in the first place. Getting some downvotes (which nevertheless I will remove right now, for you to feel happy) is nothing compared to calling me a monkey or other things.

@shintuku I didn't talk with you at the first place. I thought Ted was ignoring and I sent him a message. You jumped in the middle from nowhere talking about freedom and the ways of the world.

@TedShifrin So it seems your definition of derivative is different from what is in my mind, and perhaps that's the root of confusion. What you mean by derivative of the curve is having tangent line at each point with slope dy/dx regardless of parametrization?

@TedShifrin ok I will read that.

I mean taking simply the derivative of the components at each point. This is how Pugh defines derivative; from the domain to the Linear transformations, and each linear transformation is represented by a matrix (Here it is a 1x2 matrix)

@TedShifrin Pardon for the bad usage of the term "two variable". I mean its image is in R^2 and the domain is [0,1]. I look the curve as a function of t and take its derivative with regard to any t in the domain. Let me emphasise that this is not necessarily the same as the slope of tangent line. Like the circle (sin(t^2),cos(t^2)) that has tangent line everywhere but (in my definition of) derivative has a (0,0) derivative at t=0.

@shintuku Take a look at the link. Definition 2.1 (c). I am free to tag or not. #Respect

Take a look at this for the definition of regularity (compatible with what I mean the derivative to be): maths.dur.ac.uk/users/pavel.tumarkin/past/fall16/DG/…

@shintuku Take a look at [Pugh, Chapter 5, Theorem 10]. What I mean by derivative is not dy/dx, I mean the derivative of a function of two variables, with the explanations I told you above. I don't care what you mean by derivative. Regularity means the derivative of all of the components can not be zero at the same time. #Respect

@shintuku Read his answer. His answer works only for curves with non-zero derivative. But (sin(t^2),cos(t^2)) has a zero derivative at t = 0. Doesn't it violate the assumption of regularity? But yes, if we change the meaning of "derivative" to having a tangent line at each point, perhaps "your" meaning of derivative, then yes, his answer works because the image remains the same under 2 parametrization.

I mean for example (sin(t),cos(t)) has an image of a circle. Also (sin(t^2),cos(t^2)) has the same image. Your solution works for the first parametrization, because it's regular, but doesn't work for the second, because it has derivate (0,0) at t=0. But nevertheless we can CONSIDER the circle as a union of 2 smooth arcs, either with the first parametrization or with the second. But your solution doesn't work for the 2nd parametrization. That's where I get confused.

@TedShifrin Oh, I see. But are we not trying to change the shape of the curve such that it can not be written as finitely many arcs? Do you mean that the image of the curve in R^2 with one parametrization is regular, and by the theorem can be split it into finite smooth arcs, and then by changing the parametrization the same can't be done although the image doesn't change? Did I understand correctly or I am deeply confused? XD

But it seems I am wrong.

@TedShifrin OK I will think about your new suggestion. I thought these cusps where the semi-circles meet don't harm smoothness (with my definition of being of a class C infinity), just like the cusp in (t^3,t^2) at t=0 doesn't harm smoothness, and it has zero derivative there.

@TedShifrin Good suggestion. I have an example in my mind. For each interval in the x-axis like [1/n, 1/(n+1)], I define my curve as a semi-circle (drawn above the x-axis) with the diameter equal to the length of the interval. The semi-circles keep shrinking up to zero. At the endpoints (where x = 1/n) the curve has cusps, but I think it is also smooth there (not sure about that) because it is where the two tangent semicircles meet . Do you think it works?

Or, in other words, I don't know whether an smooth parametrization exists for the image I have in my mind (for example the image of the sawtooth function). That's where I feel lost. And yeah, I feel I am spending more energy than required, out of curiosity XD. And I feel you are right about the intention of Pugh.

@TedShifrin Thanks for the response. I thought you were done with my approach or didn't like it :DD. Yeah, the issue is putting together an infinite number of curves with cusps. I've understood until here. I mean I can imagine such curves, by reading your examples. But my problem is I am not sure how to make sure that those curves are smooth, by my definition of smoothness.

An interesting case is the curve (a simple circle) defined by (sin(t), cos(t)). When we take the derivative, we see that it is equal to (cos(t), -sin(t)) it is nowhere equal to (0,0). But for another curve with the same image, but a different parametrization, defined by (sin(t^2), cos(t^2)), we have that the derivative is zero at t = 0. So if we change the parametrization it becomes non-regular, If I didn't miss anything.

and (t^3,t^2) is a classical example of a curve that has a cusp, but is nonetheless differentiable at t=0. After reading the book, it may be a good exercise to read some classical examples too, as it helps for beginners.

*p is a point in the interval

Suppose we have a multivariable function f (here a parametric curve) defined from [0,1] to R^2. The derivative of the curve exists at p if and only if each of its components are differentiable at p. [Pugh, Real Mathematical Analysis, Chapter 5, Theorem 10]. The level of the book is higher than the level of 10th grade math, by the way, so one needs preparation to understand it.

Respect is more important than curves

@shintuku I think it doesn't need insistence from my part to show that I prefer not to talk with you. But you can continue as you like.

@shintuku I am talking to myself now. Or you are talking to me?

@shintuku Yeah, and this is the message where I started, with a post-modern meaning of "I", perhaps an intersubjective "I".

@shintuku You need to take Advanced calculus or read about it to understand the meaning of multivariable derivative, then we can talk, and of course I am free, like Ted, to answer your insult or not. I prefer not. Best wishes.

@shintuku No need to interfere when you don't know the basics of the issue.

Derivative of this function is defined from [0,1] to L(R,R^2). At each point the derivative is a linear transformation, represented by a matrix, and this matrix is usually called the "derivative".