cyrille.piatecki
• Member for 5 years, 10 months
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I hope this Clear[A, pts1, pts2, pts] A = {{.80, 0}, {0, .64}}; pts1 = Table[NestList[A.# &, {k, 3}, 20], {k, -3, 3}]; pts2 = Table[NestList[A.# &, {k, -3}, 20], {k, -3, 3}]; p1 = ...

I think that what I was expecting is this centering := CellPrint[ExpressionCell[#, "Output", TextAlignment -> Center]] & and Plot[x^2,{i, 1, 10}]// centering

Since there does not seem to have a [UpTee] here is a ugly solution ut = Row[{Rotate[Style["\[DoubleRightTee]", 20, Bold], 120.95]}]; Row[{Style["x "], Style[ut] , Style[" y"]}] Hope it will be ...

Mathematica say to use reduce so use Reduce instead Reduce[y < -3 + 2 x && y >= 1 - x]

I am badly placed to critic you question since I use to be as criptic as you. First only an economist can understand your objective since this representation is not so standard as you think. I have ...

Suppose the number of data are a multiple of 4. This is a simple way m = HoldForm[{1 2 3 4, 5 6 7 8, 9 10 11 12}] /. a_ b_ c_ d_ -> {a, b, c, d} Try m//MatrixForm this doesn't work since ...

This is the easiest way to solve the individual problem obj := p Log[c] + Log[d] + Log[l] constraint := (1 - t) w (1 - l) + T - c - (1 + τ) d log = obj - λ constraint grad = Grad[log, {c, d, l, λ}] ...

In your case you can eliminate simply the duplicate by DeleteDuplicates[list[[1]]] but you perhaps this may be a part of a huge procedure and you do not know which numbers are duplicates. This is ...

Ok it may be strange to put an assumption on t but this A[1][t_] = Piecewise[{{1 - 2*t, 0 <= t <= 1/2}}, 0] Assuming[0 <= t <= 1, Integrate[A[1][x], {x, 0, t}]] works

Inspiring by an old question I asked, here is a possible way to encode your graph ClearAll[esF] esF[setback_: .2, as_: .1] := {Arrowheads[as], Arrow[GraphElementData[{"CurvedArc", "Curvature"...

Since I must leave and I have no response, I set a very simple example: f[x_,theta_] := Sin[x^2]/(2 - theta) Plot[Thread[f[x, Range[0, 1.5`, 0.05]]], {x, 0, 3}] Obviously, this question is a ...

Here is a simple answer Define k[n] as a listable function : SetAttributes[k, Listable] k[n_] := \!$$\*SubsuperscriptBox[\(\[Integral]$$, $$3$$, $$10$$]$$\*FractionBox[\(1 + Cos[f\ \((1 + 2\ n\ ... View answer Accepted answer 3 votes As a beginner and certainly an economist you should be aware of two points : -- in Mathematica you should avoid capital letters as variable because capital letters are for build-in commands. And pay ... View answer Accepted answer 3 votes Could this be satisfacing ? Graph[{1 -> 2, 1 -> 3, 1 -> 4, 2 -> 5, 3 -> 5, 3 -> 6, 4 -> 6, 5 -> 7, 6 -> 7}, VertexCoordinates -> {{0, 1}, {1.1, 2}, {1, 1}, {1, 0}, {... View answer 3 votes Try this f = Sin[#] & der = D[#, x] & der[f[x]] or this der[f_[x]] = d[f[x], x] l[x_] = a x + b x^2 der[l[x]] Normally your definition should works --- it works on my computer. Restart the ... View answer Accepted answer 3 votes Plot[{50 p - 600, -25 p + 1200}, {p, -10, 10}, PlotLegends -> Placed["Expressions", Below]] or Plot[{50 p - 600, -25 p + 1200}, {p, -10, 10}, PlotLegends -> Placed[{"f1", "g1"}, Below]] View answer 3 votes Sorry, I will not give explanation because I am in hurry m = Table[Subscript[a, i, j], {i, 1, 4}, {j, 1, 4}] m // MatrixForm Flatten[m] l = Partition[m, {2, 2}] l // MatrixForm In the same ... View answer Accepted answer 3 votes Your operation does not work because in this case it is not a matricial multiplication but a composition. Try mM = {{D[#, x] &, D[#, y] &}, {D[#, y] &, D[#, z] &}} v = {f[x, y, z], ... View answer 2 votes Manipulate[ Plot[a + b*x, {x, 1, 100}], {a, -1, 10, 1}, {b, -10, 10, 1}] View answer 2 votes The MMM answer is the simplest but remember how in textbooks they show how to solve an exact differential equation : Solve[\!\( \*SubsuperscriptBox[\(∫$$, $$a$$, $$x$$]$$s \[DifferentialD]s$$\) == \!\...

There is always many way to arrive to the same solution. As the last commer I give the doc way --- have a look to http://reference.wolfram.com/language/tutorial/Differentiation.html S10 = x^2 E^(a y -...

Here is an other approach but still with bad results --- I have looked some years ago to the estimation of this kind of sde and I think it is normal step := 0.01 tt := 120 data = RandomFunction[ ...

The only problem is that for a $n \times 1$ matrix, Dimensions may b a singleton so g = If[MatrixQ[#] && Length[Dimensions[#]] == 2 && Dimensions[#][[1]] == Dimensions[#][[2]], ...

Perhaps simply this t=Table[{If[i == 1, Missing[NotApplicable], i], j}, {i, 1, 2}, {j, 1, 2}] Drop[t, 1, None] // TableForm

I hope this will be the expected answer Clear[taxicab, x1, x2, y1, y2]; taxicab[{{x1_, y1_}, {x2_, y2_}}] := Abs[(x2 - x1)] + Abs[(y2 - y1)]; taxicab[{{2, 3}, {1, 4}}] l = {{0, 0}, {1, 0}, {1, 1},...

First, of all people in MSE need data and/or code to try to help you. So be always aware to ask a question with gives the minimum effort to the one who will try to answer --- I have use an image ...

I am OK with Chuy but the best thing is to look at code f[x_] := Sin[2 x] g[x_] := 2*Sin[x] Plot[{f[x], g[x]}, {x, -1, 1}] every command in mathematica begins with an uppercase. Forget ( for the ...

Unfortunately, you have not documented the values of the parameters. As for the Lorentz 3D equation, there is no closed form solution. Here is the simple code for which you asked --- obviously, your ...

If I understand your question, there is no need of Mathematica to solve your problem for $x_0$ given, we know that $y_{n+1} = y_{n} + 5$. We know also that $y_0 y_1 = 12500$. That is to say that \$y_0 ...