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3

Here n is associated to the number of starts rho is associated to the lead screw external radius delphi is associated to the tooth thickness lambda is associated to the lead n = 4; rho = 0.5; delphi = 0.3; lead = 1; lambda = lead/(2 Pi); colors = {Blue, Red, Green, Yellow, Brown, Black, Cyan, Orange}; gr0 = Table[ParametricPlot3D[{rho Cos[...


5

I've been working with multivalued functions for a while and without a doubt the best way I have found to avoid branch cuts, in particular when integrating over multivalued functions, is to construct an analytically-continuous version of the function over the desired path. For algebraic functions, this is easy: convert it to its algebraic form. For ...


3

When I asked this question I was more optimistic about this problem. But it turned out this is a really hard problem. Avoid doing modulo 2𝜋 operation didn't work. It would work when an expression is made only by multiplication and powers, but addition combines two angles in a complicated way and it makes this approach difficult. So my original idea failed. ...


1

@Hubble07 has been criticized for the estimate in the comment. However, I do believe it provides a valuable analytic insight that can be improved a little bit to provide the bounds. The area of the $R$-sphere covered with one layer of $r$-spheres reads $$A=4\pi(R+r)^2.$$ The area per small sphere is within the areas of hexagon and triangle in which it is ...


4

Reversing the graph and reformatting the nodes should get you started: ClearAll[hasseF] vf[{xc_, yc_}, name_, {w_, h_}] := Text[Grid[name, Dividers -> {False, True}], {xc, yc}]; hasseF = ReverseGraph@*TransitiveReductionGraph@*RelationGraph hasseF[ SubsetQ, Subsets[{{M1, W1}, {M1, W2}, {M2, W1}, {M2, W2}}], VertexShapeFunction -> vf ] Edit: @...


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