So identity means whatever happens to an element, you get that element back.
I think I have a math theory
- Thread starter Vast3
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yeah.
elements within a curve possess contrariness when each element has a corresponding inverse. For a curve C and operation ⊕, for every P ∈ C, there exists an element P′ ∈ C such that:[math]P \oplus P' = P' \oplus P = e[/math] where e is the identity element.
2.6 Curvature.
okay but if you read the pdf you can see all of these, Curvature defines how a curve deviates from being straight or flat. It captures the degree of bend or deviation of the curve’s path from a linear trajectory. For parametric curve γ(t), the curvature κ is given by:[math]\kappa = \frac{|\gamma'(t) \times \gamma''(t)|}{|\gamma'(t)|^3},[/math] where γ′(t) and γ′′(t) are the first and second derivatives of γ(t) with respect to t.
I have seen this formula before, but I don't remember where. Can you prove that a straight line has a zero curvature by this formula?
To prove that a straight line has zero curvature, we use the definition of curvature for a curve in one dimension (for lines in one dimension):Let [imath]y = mx + c[/imath] be a straight line defined by the linear equation [imath]y = mx + c[/imath], where [imath]m[/imath] is the slope of the line and [imath]c[/imath] is the y-intercept.1. \textbf{First Derivative}: The first derivative of this line with respect to [imath]x[/imath] is [imath]m[/imath], meaning: [math]\frac{dy}{dx} = m[/math]2. \textbf{Second Derivative}: The second derivative of the line with respect to [imath]x[/imath] is zero, since the first derivative [imath]m[/imath] is constant and does not change: [math]\frac{d^2y}{dx^2} = 0[/math]3. \textbf{Curvature}: Curvature of a curve in one dimension is defined by the second derivative being non-zero. Since [imath]\frac{d^2y}{dx^2} = 0[/imath] for the straight line, it implies that the curvature is zero. Therefore, the straight line [imath]y = mx + c[/imath] has zero curvature because its second derivative with respect to [imath]x[/imath] is zero throughout.
Do you know what is the difference between you and me?
what ?
I can answer the question in one sentence. If the curve [imath]\gamma(t)[/imath] represents the straight line, it is so obvious that [imath]\gamma''(t) = 0[/imath], so the curvature [imath]\kappa = 0[/imath]. I have answered it in a short sentence because I have never let AI solve my problems. Even If I struggle in a problem, I don't let it solve it, instead I bring it here in this forum to discuss its solution.
Tell me frankly, did you use an AI to solve the problem?
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no, but I just used a long description because I thought you couldn't understand it that way.
But this is a contradiction to your theory. If your theory defines the curve as [imath]\gamma(t)[/imath] and you use [imath]y = mx + c[/imath], then you are not following your definitions. This only tells me that an AI has solved it for you!
yeah, this is because I desired to do it in this way but I said lets do that way and this is the long description on using my formula :
To prove that straight lines have zero curvature, we use the curvature formula for a curve:
[math]\kappa = \frac{|\gamma''(x)|}{(1 + (\gamma'(x))^2)^{3/2}}[/math]
where:
[imath]\gamma(x)[/imath] represents the equation of the curve.
[imath]\gamma'(x)[/imath] denotes the first derivative of the curve.
[imath]\gamma''(x)[/imath] denotes the second derivative of the curve.
Now, consider a straight line [imath]y = mx + c[/imath]:
First Derivative: The first derivative of the straight line [imath]y = mx + c[/imath] is constant:
[math]\gamma'(x) = m[/math]
Second Derivative: The second derivative of the straight line [imath]y = mx + c[/imath] is zero, as the first derivative [imath]m[/imath] is constant:
[math]\gamma''(x) = 0[/math]
Curvature Calculation: Substituting the values of the second derivative and the first derivative into the curvature formula gives:
[math]\kappa = \frac{|\gamma''(x)|}{(1 + (\gamma'(x))^2)^{3/2}} = \frac{0}{(1 + m^2)^{3/2}} = 0[/math]
Therefore, straight lines, such as [imath]y = mx + c[/imath], have curvature zero, indicating they do not bend in any direction.
Fantastic. It seems that you understand more than me in curves. Go to the new topics posted today and yesterday, you will see this title:
Differential geometry and vector fields exercise
I think that you can be very helpful to give him some hints and advices to solve the problem he posted.
okay, but there is hypothesis in my theory that I can't solve, in PDF you can see it.
What is the hypothesis?
this is the hypothesis: Suppose there exists a curve C with a root point t. The hypothesis states that there are points near t that generate a region where the curve does not open up. Formally, for a curve C and root point t, there exists a neighborhood U around t such that for any point p ∈ U, the curve C remains closed within this neighborhood.
I have not learnt something like this in calculus when I studied Curves. My knowledge in this topic is very limited. Curves outside the world of calculus, I can understand some of their notation (But really cannot solve them). Like the title I have sent to you. I understand what is written there and most of the notations I know them, but I cannot help in solving the problem.
My advice for you is to make a new thread and post this problem there. Professor blamocur or professor Dave might be able to help you solve this problem or hypothesis or whatever you will call it.
I have really enjoyed learning some new things with you today and I hope that one day you will post a problem in curves that I fully understand, so that I can help you in solving it.
Cheers,
-Mario, Dan's student
