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| It's towards the bottom of the Details and Options section in the documentation for `MeshRegion` and `BoundaryMeshRegion`. ![enter image description here][1] [1]:... |
| You can choose to sample many points as `t` varies and few points as `u` varies. Choosing `PlotPoints -> {2000, 3}` gives good detail: r = 7; R = 19; ParametricPlot3D[{{(u^Cos[(r t)/R]) Cos[t], (u^Cos[(r t)/R]) Sin[t], (-Log[u] Sin[(r... |
| You can get around 1.5 times faster if you avoid the pattern matcher by replacing the lines containing `ReplacePart` with the following: Change ajm1G = ReplacePart[ajmG, {i_, i_} -> 1]; ajm1H = ReplacePart[ajmH, {i_, i_} -> 1]; to ... |
| Ah, ok! When deriving the new domain, I had a typo and used $t = e^{-x}$ instead of $x = e^{-t}$... |
| A friend of mine approached me with the problem of wanting to break up a Graphics object like so: ![enter image description here][1] The only solution I could think of involved turning the Graphics object into a Graph, where points become... |
| Hi Joseph, I guess I'm a bit confused. In your first diagram, how can you infer that the upper 2 and 3 means 2^3 and not 2*3, 2+3, etc? Also how does your idea scale with arbitrary functions? For example can you make a diagram for y = sin(2x)? ... |
| The numerator in the fractional exponent was being overlooked, so you were seeing the root plot for i^(1/7) instead of i^(2/7). |
| I've often found exporting to eps works better, then you can convert that to pdf. Perhaps this will work for you. |
| Expanding on that idea, we can take real and imaginary parts: re = Refine[ComplexExpand[Re[I^n (n^(1/n) - I)]], n > 0] (* n^(1/n) Cos[(n ?)/2] + Sin[(n ?)/2] *) im = Refine[ComplexExpand[Im[I^n (n^(1/n) - I)]], n > 0] (*... |
| I'm not sure how the primitive root is chosen, but here are some ways to find the smallest primitive root in Mathematica. In Mathematica 10 you can simply do First[PrimitiveRootList[n]] In Mathematica 9 we need to define a new function ... |