E. Chan-López
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Try this: Simplify[Refine[Asymptotic[Sech[a x], a x -> ∞], Assumptions -> Element[a,Reals]]]

Try with Reduce[] command: Reduce[(1 + x + x^2 + y - x y + y^2) == 0, {x, y}, Reals] (*x == -1 && y == -1*) Solve[] works too: Solve[(1 + x + x^2 + y - x y + y^2) == 0, {x, y}, Reals] (*x == -...

Try this: p[x_,y_, a_] := 7 + 2 a (x + 1)^2 + 3 a^4 (y - 2) (x + 5) Collect[a*p[x, y, a], a, FullSimplify]

Other way: For your graphics labels, use: << MaTeX` Definition of the $\phi(x,n)$ function: a = 1; p[f_] := Plot[f, {x, 0, 1}, PlotRange -> {{-0.08, 1.08}, {-60, 280}}, PlotStyle -> {Blue,...

I recommend that you review the Liu criterion for the Hopf bifurcation, since it is the most suitable for detecting limit cycles in systems with three or more equations. It is more convenient to write ...

Try this: twoEqs = {d + (b + a c) x + (a + b + c) x^2 == 0, e + (b + a f) x + (a - c + k) x^2 == 0} Reduce[CoefficientList[twoEqs[[1, 1]], x] == CoefficientList[twoEqs[[2, 1]], x]] (*(d == e &&...

First, try registering the external evaluator (Python or other). You just have to locate the path where you have it installed. Test:

$Version (*12.3.0 for Microsoft Windows (64-bit) (May 10, 2021)*) Try this: dots[x_, y_, z_] := {{MaterialShading["Plastic"]}, EdgeForm[None], Sphere[{x, y, z}, 0.105], Lighting -> &... View answer 1 votes Try this: list = {};(*empty list*) Do[AppendTo[list, Det[Table[(-0.15 + 0.001*x)^(i + j), {i, 1, 9}, {j, 1, 9}]]], {x, 1,10}] Select[list, # > 0 &] (*{6.89637*10^-201, 6.10433*10^-201, 2.15982*... View answer Accepted answer 1 votes Try this: t = {}; Do[f[z_] := Sin[z]; Sol1 = {z, z f[z]}; AppendTo[t, Re[Sol1]];, {z, -4, 4, 3/4}, {p, {10}}]; PrependTo[t, #] &@{"z", "z f[z]"} // Grid[#, Alignment -> {&... View answer 1 votes Try: Export["rotation-debug2.gif", Rasterize[img, RasterSize -> 700, ImageResolution -> 1000]] View answer 1 votes You can use a Table: With[{x0 = {1, 3, 6}, t0 = 0,tfin = 5}, solx[s_] = Table[NDSolveValue[{x'[t] == -x[t], x[0] == x0[[i]]}, x[s], {t, t0, tfin}], {i, 1, Length[x0]}]; Plot[Evaluate[solx[t]], {t, 0,... View answer 1 votes Try with the following code: LieBracket[field1_?VectorQ, field2_?VectorQ, vars_?VectorQ] := Module[{jac, lieb}, jac[field_?VectorQ, vars1_?VectorQ] := D[field, {vars}]; lieb = jac[field2, vars].... View answer 1 votes Try with the following code: RowsSum[nmax_Integer?Positive, length_Integer?Positive, vector_List] := Module[ {matrix, matrixrows, s}, matrix = Table[i, {i, nmax}, {length}]; matrixrows[... View answer 1 votes According to the documentation: (*Recover a function from its gradient vector*) sol = DSolve[{D[x[u, v], u] == Cos[u - v] Cos[u - v], D[x[u, v], v] == Sin[u - v] D[Cos[u - v], v]}, x, {u, v}][[1]]... View answer 1 votes Your system: f1[y1_[t_],y2_[t_]]=4*y2[t]+y1[t]; f2[y1_[t_],y2_[t_]]=y2[t]*y1[t]+y1[t]-y2[t]; F[{y1_[t_],y2_[t_]}]:=Evaluate[{f1[y1[t],y2[t]],f2[y1[t],y2[t]]}]; X={y1[t],y2[t]}; Equilibrium points: ... View answer 1 votes Your nonlinear system is a good example for showing Hopf and Bogdanov-Takens bifurcations. With the following changes A=A0+y0;B=A0*y0; we obtain the equilibria: (*{{x -> 1/A0, y -> A0}, {x ->... View answer 1 votes Try the following trick: realPartRule = Complex[re_, im_] :> Complex[re, 0]; realPart[exp__] := exp /. realPartRule; Applying this trick to your result we obtain: realPart[Integrate[Exp[a*(x^3)], {... View answer 1 votes Use the following code: Manipulate[sol = NDSolve[{x'[t] == x[t] (1 - x[t]/7) - (6 x[t]*y[t])/(7 (1 + x[t])), y'[t] == 1/5 y[t] (1 - n y[t]/x[t]), x[0] == 1, y[0] == 1}, {x, y}, {t, 0, 200}]; Grid[{{... View answer 1 votes Try with the package CurvesGraphics6 (for more details see Gianluca Gorni): The function$V(x,y)$: V[x_, y_] := Exp[d x + b y]/(x^c y^a) /. {a -> 18/10, b -> 9/10, c -> 81/100, d -> 54/100}... View answer 1 votes Try this: f[t_] := fp(t);(*Cos[t]^2*) n = 26; tm = N[Range[n - 1]]/n; fm = Map[f, tm]; coefs = FourierDST[fm, 1]/Sqrt[n/2] coefc = FourierDCT[fm, 1]/Sqrt[n/2] View answer 1 votes Contributing to Chris's analysis: Taking$q=\displaystyle\frac{33}{4}$,$r=\displaystyle\frac{1}{90}$,$\alpha_{1}=\displaystyle\frac{64}{33}$,$\alpha_{1}=\displaystyle\frac{1}{22}$,$\beta_{1}=24$,$...

Try this: Collect[D[k[θ[t]]*θ[t] - θ'[t], t] == 0, {D[θ[t], t], D[θ[t], {t, 2}]}]

Try this: With[{a = e d, b = f d,c = g d}, eq = a*x + b*y + c*z == d; Map[PolynomialQuotient[#, eq[[2]], x] &, eq]] (*e x + f y + g z == 1*)

Try this: Table[Plot[Sqrt[2/Pi] Sin[(n Pi x)/Pi], {x, 0, Pi}, AxesLabel -> {"x", "\[CapitalPhi]"}, PlotLegends -> {"n = 1", "n = 2", "n = 3", &...

Your finite sum exists using a conditional (If): Simplify[Sum[If[i === j, 0, (a - Subscript[a, i])/(Subscript[a, i] - Subscript[a, j])], {i, 0, n}, {j, 0, n}]] (*-(1/2) (1 + n)^2*)

Another option: fcompiled = Quiet[Compile[{x}, 1 + Cos[x], CompilationTarget -> "C"]] f[x_?NumericQ] := If[x \[Element] Reals, fcompiled[x], HoldForm[f[x]]] Tests: (* Test 1: *) f[I] (*...