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| I wonder if it is surprising that having 11 parameters (even though restricted to integers) is able to predict just 9 numbers. I think there are an infinite number of solutions in the form heegner = Sum[a[i] n^i, {i, 0, 7}]/Sum[b[i] n^i,... |
| Apparently *Mathematica* needs to know explicitly about each version of R. Look at the *Mathematica* folder for your operating system which for me is C:\Program Files\Wolfram... |
| It's more than odd. The two approaches you give should give the same result but only the second one below does give the right answer. (You should inform Wolfram, Inc.) FullSimplify[Probability[x y >= 0, {x, y} \[Distributed]... |
| For whatever it's worth the second answer followed what's done for a "wrapped normal" distribution. For the first answer, I chose a 1-to-1 formula and `Abs[xx]` is not 1-to-1. |
| Changing the starting value from 0.01 to 0.98 works. |
| Because `c` and `m` only occur as `c^m`, you can only deal with `c^m` and can't get separate estimates for `c` and `m`. Also if you change `0.75` to `3/4` and set `t` to specific values, a solution for `c^m`, `a`, and `b` occurs. t = 4; ... |
| Are you wanting *Mathematica* to return something like Piecewise[{{(x^-n (-1 + x^n))/(-1 + x), x != 1}, {n, x == 1}}] ![Piecewise formula][1] [1]:... |
| If you consider a truncated Poisson (i.e., getting rid of 0), then *Mathematica* can do all of the heavy lifting. First we find the distribution of a truncated Poisson: d1 = TruncatedDistribution[{1, \[Infinity]},... |
| I agree. The dataset seems contrived as it is perfectly symmetric about -1/2: data[[1 ;; 20, 2]] == -data[[Range[40, 21, -1], 2]] (* True *) ListLogPlot[Transpose[{data[[All, 1]], Abs[data[[All, 2]]]}]] ![Plot of data on a... |
| Your triple integration is Integrate[xy*sqrt(x^2+y^2+z^2),Element[{x,y,z}, ImplicitRegion[z>=sqrt(x^2+y^2)&&x^2+y^2+z^2 Sqrt[x^2 + y^2], x^2 + y^2 + z^2 |