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| That's great, it could win a code-golf competition. BTW, I had posted this as a reply to a question, but now the question appears to have disappeared - was it deleted by the user? |
| That's great! I cannot count the number of times I've spent hours on something only to find out it is done by a built in function lol. I worry that all the cool stuff has already been coded before I start work there.... |
| Use a `ParametricPlot` instead, ParametricPlot[{{x, x^x}, {x^x, x}}, {x, -1, 3}, PlotRange -> {-1, 3}] ParametricPlot[{{Sin[x], x}}, {x, -3, 3}, PlotRange -> {-3, 3}] ![Mathematica... |
| `GraphPlot` will work when given an `AdjacencyMatrix`, but if you look at the structure of `yy` then you see it isn't **quite** a matrix: In[5]:= Dimensions[yy] Out[5]= {1, 10, 10} You can see that `yy` is a list of matrices, a... |
| Yeah, that Macau really messed up the curve, is why I needed a log scale. I guess I just expected the `PlotRange` and `ColorFunction` to behave like they do in `DensityPlot` or similar. But it doesn't quite, and I can't really find the rhyme or... |
| You can look into the various ways of plotting multivariable functions, look [here](http://mathematica.stackexchange.com/q/64674/9490) and [here](http://mathematica.stackexchange.com/q/25277/9490). To get a plot larger, use the `ImageSize` option.... |
| Move both of the `Epilog` points to be part of the `Show` rather than the individual plots Manipulate[ Show[ Plot[x^2, {x, 1.`, 10.`}, PlotRange -> Automatic], Plot[1.5 x^2, {x, 1.`, 10.`}, PlotRange -> Automatic] ... |
| It doesn't work for the same reason that ClearAll[a, b, x, f] f[x_] := (a x^2)/(b + x) With[{a = 10, b = 5}, Plot[f[x], {x, -10, 10}, PlotRange -> {-20, 20}] ] ![Mathematica... |
| ![enter image description here][1] &[Wolfram Notebook][2] [older image]: https://community.wolfram.com//c/portal/getImageAttachment?filename=VisualizingInterstellar%27sWormholefromarticletoprogramming.gif&userId=20103 [1]:... |
| Your integral diverges at the limit of integration. When `theta = ArcCot[Cos[phi]]` then the denominator of the integrand is zero. |