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For $P=(0,0)$ and $F=Y^2-X^2-X^3, \; G=Y+X,\;H=Y-X$ we get $$I(P,F\cap (G+H))=2\quad \text{and} \quad I(P,F\cap G)=I(P,F\cap H))=3$$ and indeed $2\geq \min(3,3)$ is not true.


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This is in general wrong. The curve $[0,2\pi]\ni t\mapsto (\sin (\tfrac{t}{2})\cos (\tfrac{t}{2}),\sin (\tfrac{t}{2}))$ provides a counter-example.


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Being closed has nothing to do with it. At any sharp point on the curve, the curve traced by the derivative will be discontinuous. The start/end point is not in any way special, it's just a matter of having a parametric curve with a continuous derivative. You could have a closed curve that is continuously differentiable at start/end point, but has a cusp ...


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It sounds like you are describing something called a Triangle Wave. There are several different ways to write it. One of the easiest to describe would be using piecewise defined using the fractional part of $x$ (defined as $x - \lfloor x \rfloor$) Other possible patterns include square waves and sawtooth waves, depending on what it is you want. Play ...


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Hint. The uniformizing parameter of $\mathcal O=k[X,Y]_{(X,Y)}/(F)$ is $y$ (the residue class of $Y$ modulo $(F)$).


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It means a small change in the value of function $f$. You slice up your function curve in infinite possible elements, and then each of will be $df$ in length, and integrating w.r.t. $df$ or $dy$ then means you are summing up all those small slices. This way we can find area under a curve or length of a curve using double or single integrals, also volume ...


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$$ \gamma'(t) = (-\sin(t) + 1.5 \sin(3t), \cos(t)+7\cos(7t)+3\cos(3t))\\ \| \gamma'(t) \| = \sqrt{(-\sin(t) + 1.5 \sin(3t))^2, (\cos(t)+7\cos(7t)+3\cos(3t))^2} $$ In general, for a vector $(a, b)$ in the plane, $(-b, a)$ is perpendicular to it. So your normal vector is $$ N(t) = \frac{\pm (-\cos(t)-7\cos(7t)-3\cos(3t), -\sin(t) + 1.5 \sin(3t))}{\| ...



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