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W. Craig Carter
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DatePlus is expecting a number or a list as its second argument. The function DatePlus[{2021, 1, 10}, x] is coming back unevaluated and a warning is being printed. When you replace the x, then it is being evaluated correctly. Compare:...
This will give you the Form representation of the entity: Entity["PhysicalConstant", "ElectricConstant"] This will give you the Symbol associated with it: Entity["PhysicalConstant", "ElectricConstant"]["Symbol"] You can do this: ...
Your data is a bit noisy which is normal for experimental data: yourData = Import["~/tmp/ASmappa-niccolini90-0.dat", "TSV"]; you will want to use your path to your data. dataForInterpolate = yourData /. {x_, y_, z_} :> {{x, y}, z}; ...
I made an error in my first example: There are real solutions: FindInstance[(y - x^2)^2 + (z - x^3)^2 == 0, {x, y, z}, Reals, 6] // N Solve[(y - x^2)^2 + (z - x^3)^2 == 0, z] Reduce[-x^4 + 2 x^2 y - y^2 >= 0, {x, y}, Reals] ...
Hello Alex, A bunch of choices: Caliing Exp on a list: 1000 Exp[Range[0, 1, .01]] Using table with an iterator Table[1000 Exp[-v], {v, 0, 1, .01}] Iterating over a list With[{values = Table[x, {x, 0, 1, .01}]}, ...
Here is a way to get something, but I wouldn't expect it to be useful or reliable. peaks = FindPeaks[data] Find an estimated distribution for the peaks (without the last peak): peakDist = FindDistribution[Last /@ Most[peaks]] Find...
A couple things: 1) When you do this: m = {{1,2},{3,4}}//MatrixForm m is no longer a matrix, it is a "Form" and m.{x,y} will not operate as you might expect. 2) It is much easier to debug if you put your input into separate...
When you said function in your original post, I misinterpreted what you meant: The two expressions funcPattern[x_] := x^2 + 1 and function = Function[{x}, x^2 + 1] have the same behavior to the user, but they are different...
The first argument of Module (or Block) is a list of variables that you want to "limit the scope to the Module (or Block))" For your example: Define i i = Green Define your functions (localizing i in the second is superfluous, because For...
Your "inner" integrand: (R - r*x)*(3 x^2 - 1)*E^(-1.5566*b) evaluates to an expression that depends on theta and r. You probably want to integrate over theta. Something like (you will want to check this assumption so that it corresponds to...