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The concept of angle is not going to be well-defined for such curves: consider $ y = x $ and $ y = \lvert x \rvert $. What is the angle between these curves at $(x,y)=(0,0)$? It could be $0$, approaching from positive $x$, or $\pi/2$, approaching from negative $x$. And that's the least pathological case, where the left- and right-hand derivatives exist (and, ...


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No. [N.B. the first part of this answer addresses the original version of the question where the assumption $\kappa \neq 0$ is made.] The trivial counterexample: most people use the word "convex" in the "closed sense", in which a straight line is a convex curve. A straight line does not have non-vanishing curvature. Assuming that you actually meant ...


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We may use a variation of Laplace's method: let $f_n(x)=\sqrt{1+n^2(1-x)^{2n-2}}$. $f_n(x)$ decays pretty fast on $[0,1]$, and by considering the Taylor series in a right neighbourhood of the origin we have: $$ f_n(x)\approx \max\left(1,\sqrt{1+n^2}\,e^{-nx}\right)\tag{1} $$ so that: $$ l_n \approx 1+\frac{1}{n}\sqrt{1+n^2} \tag{2} $$ gives that the limit ...


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Matrix derivation The original equation can be described in a more concise form as $$\vec x A \vec x^T + \vec x \vec b^T + \vec b \vec x^T + f =0$$ or $$\vec x A \vec x^T + 2\vec x \vec b^T + f =0$$ where $\vec b=(d,e)$ and $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$. If the vector $\vec x_0$ represents the center then in a coordinate system with ...


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I would emphasize that a pure translation in the plane preserves the relationship between a figure and its center. Given $$ A x^2 + 2 B xy + C y^2 + 2 D x + 2 E y + F = 0 $$ and solving for $(x_0, y_0)$ in $$ A x_0 + B y_0 + D = 0, $$ $$ B x_0 + C y_0 + E = 0, $$ we introduce translated coordinates $(u,v)$ with $$ x = u + x_0, $$ $$ y = v + y_0. $$ I'm ...


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If the conclusion does not hold, place a fifth point in the component that does not contain the curves C1 and C2. Connect this fifth point to the four points X, Y, Z, W by connected arcs that do not intersect each other or meet the Jordan curve or cross it. The result is a complete graph $K_5$ on five vertices in the plane with no self intersections. This is ...


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I put three in a chain: I use the positive domain of the NTS to 'skew' the input, the second is the basic sigmoid, and then I use the positive domain to skew the output. The result is an extremely flexible curve which I tend to use everywhere now. Here is a Desmos sheet, I think it is what you were looking for, although the other answers seem good too. ...



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