New answers tagged

0

Oddly, technical support at $Mathematica$ was unable to reproduce the issue, though @Xavier and I were both able. They asked me to try a clean start, but the crash still occurred. I've sent my SystemInformation[], and this issue is being tracked under CASE:3588559. The technician's last comments: I have filed a report with our developers which includes ...


12

The question title poses a good question, although the question formulation is somewhat specialized and misplaced (as mentioned in a comment). This answer provides data and a method description answering: How to find outliers in 3D numerical data? Data It order to provide a good answer it would be better to use "real life" data. Not spending much ...


3

$Version (* "10.4.1 for Mac OS X x86 (64-bit) (April 11, 2016)" *) Hex[exp_] := FromDigits[exp, 16]; LByte[exp_] := BitAnd[exp, Hex@"00ff"]; HByte[exp_] := BitAnd[exp, Hex@"ff00"]~BitShiftRight~8; PRNG[v_] := Module[{L5, H5, v1, v2, carry}, L5 = LByte@v*5; H5 = HByte@v*5; v1 = LByte@H5 + HByte@L5 + 1; carry = HByte@v1~BitGet~0; v2 = ...


2

Clear[x]; x[0] = 0; T[x_] := Piecewise[{{1 - x, 0 <= x < 1/7}, {(x + 6)/7, 1/7 <= x <= 1}}]; a[n_] := n/(n + 1); b[n_] := n/(n + 5); Include memorization (Functions That Remember Values They Have Found) in the definition of x for effciency x[n_] := x[n] = (1 - a[n - 1]) x[n - 1] + a[n - 1]*T[(1 - b[n - 1]) x[n - 1] + b[n - 1] T[x[n - ...


0

T[x_] := Piecewise[{{1 - x, 0 <= x < 1/7}, {(x + 6)/7, 1/7 <= x <= 1}}]; x[n_] := (1 - a[n - 1]) x[n - 1] + a[n - 1]*T[(1 - b[n - 1]) x[n - 1] + b[n - 1] T[x[n - 1]]]; x[0] = 0.1; a[n_] := n/(n + 1); b[n_] := n/(n + 5);


0

If you were to define f with SetDelayed, then it could be used in the same way as a built-in function and all will be well. f[x_] := (-2*Exp[-x])/(Exp[x] - 2) I have made some minor changes to operation to get, what to my eye, is better visibility of the plotted graphics elements. operation[value_, f_, a_, b_, aspect_] := Block[{data}, data = ...


16

Below is given a solution derived with ILP combinatorial optimization: The total of the assigned values to the $5 \times 5$ table is $61$. I called in the comments this approach to be "brute force" because of the generation of a larger number of variables and conditions and pushing them to Maximize or LinearProgramming. Same approach was used for my answer ...



Top 50 recent answers are included