Pierre ALBARÈDE
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The link for Tensorial above is dead, you may find something interesting on J. F. Gouyet's page. I prefer in general not to use packages and make my own simple cooking. Also there are two functions ...

When OptionsPattern[] is used, OptionValue@k2 is supposed to work too, so behind the scene there is something like k2/:OptionValue@k2=3 which fails if k2 is replaced by f1[k2]: you would get a "...

The Wolfram language is more an informal abstraction than an implementation of Mathematica. A formal or normative definition does not exist yet after many years but three important events have ...

{x, y, g, a, b, c} /. {x_, y_, g_, next___} :> f1[{{x, y}, g}, f2@next] and replace f1, f2 by whatever you need.

Since the question was asked, AddSides etc *Sides have been added to Mathematica for the purpose. Simplify makes some things automatically.

Replace simplify by Simplify. To begin with, make sure that all coefficients have numerical values before launching Solve so that the solution will also be numerical. Then try again by clearing some ...

The actual output is Function[0, 999][0]. I have guessed (I hope correctly) that the expected value was 999 as from Function[a, 999][0]. In Trace[f[999], TraceOff -> Message], the essential part ...

By analogy with Python for example, I suggest that the Wolfram language be everything that can be interpreted in the command line. e. g. wolfram on the Raspberry pi or mathematica on other machines. ...

In[28]:= q_ -> # &@2 Out[28]= q_ -> 2

v12.2 update. If you have just inserted the line, then Undo will remove it. If the line is at the same height as an edge of the right bracket of a cell, then select this cell, go to menu Insert > ...

I add an answer because I need more space. Consider for example Wolfram for Software Developers and more particularly the link Wolfram language for Jupyter. The Wolfram language being passed to ...

You play a syntax puzzle. In order to see what is going on: {Sow[1]; 2; Sow[3]; 4; 5} // (Reap1 /* Last1) (* Last1[Reap1[{5}]] *) The Last of a Reap that sows nothing is obviously the empty list. {...

The normal way is simply FD1[(d - 1)/t, η] - FD1[(d - 1)/t, η - vd]/.AsymptoticSolve[eqn, {ηs, 2^(-(t/d)) nd^(t/d)}, {vd, 0, 2}]

I can answer on that: Why the hell is there no straightforward way to embed entire custom stylesheets in a notebook? Don't hack this on any important document! Open the notebook where you want ...

Try f[t_] := 0 /; 0 < t < 1 see also Step and PieceWise in the documentation.

For Newton FindRoot[{x^5 - x + 1 == 0}, {x, 1}] Then check the details with Options@FindRoot Check in the documentation the option Method in particular. For the bisection method, NSolve may use ...

In general you would use Solve but here it is straightforward: C1=0.2-xp[0] and I let you write the initial equation for C2.

In your program you can replace "a" by a to make it more general. As Mathematica is symbolic, you don't need to quote symbols. Replacing x by OptionValue@x or even OV@x in a complex expression may ...

Let me rephrase as a recipe. Do Format > Edit style sheet. The private style sheet opens: it is a notebook titled "Private style definitions for [your notebook]". In this notebook, click on the link ...

Without ListConvolve: myListConvolve[kernel_, list_][n_] := Dot[Reverse@kernel, Take[RotateLeft[list, n - 1], Length[kernel]]] example: myListConvolve[{x, y}, {a, b, c, d, e, f}] /@ Range[5] ...

The equations reduce to algebraic : {0 == (1 - x)*x + q y - x y, 0 == -q - x y + 2 f z, 0 == x - z} You can use Solve or NSolve.

"I want to get a decimal representation" is not compatible with infinite precision because some rational numbers like 1/3 have infinite decimal representation that does not fit into a ...

An example, just to summarize the main points: test = Text[ Style[#, FontSlant -> Plain, FontSize -> 32], {0, 0}] & /@ {Indexed[P, {i, j}], P} Graphics /@ test Graphics[#, ...

You get a lot of red ink. Try to simplify. Avoid complicated notations. The most useful formula: f1[x_,y_]:=f2[x,y] where f2 must not call f1 to avoid recursion (unless you want it). Don't use ...

My program is not bad but has design flaws: the Rational option is useless and the interface will fail in certain cases. So I need to update: Attributes@bisection = {HoldAll}; bisection::...

Attributes@bisection = {HoldAll}; bisection::nochangeofsign = "sign does not change, try other bounds"; Options@bisection = {AccuracyGoal -> \$MachinePrecision/2, Rational -> False}; Clear@...

Question 1 : pressure is arbitrary for any uniform shift because the gradient of uniform function is zero. There is little relation between the downstream BC and the initial conditions on pressure. ...

[Sorry this is comment again but does not fit in the box so I put answer again.] I have had a look at the tutorial and the program. There is inconsistency in your question but maybe not in the ...