# Tag Info

5

You seem to be asking for the solution to Kepler's equation for the eccentric anomaly. That is, given the mean anomaly $x$, find the eccentric anomaly $t$, where $x=t-e*Sin[t]$, and $e$ is the eccentricity. There is no closed form solution. You can use FindRoot in the following function to find t given x, a, and b. FindtGiven[x_, a_, b_] := t /. ...

5

It seems like you have the recursion t[i+2] = 2*t[i+1]-t[i]. Shifting this back two timesteps allows a standard form: t[i_] := 2 t[i - 1] - t[i - 2]; t[0] = 1; t[1] = 2; which can be solved for any i: t[#] & /@ Range[10]

4

This is not an answer but rather an extensive editing of your question. In addition to being a poor idea to use π as a variable, it is a really bad idea to use variables with subscripts as symbols as inputs to functions. This can wreak havoc and is best avoided. What I did was to replace all occurrences of a symbol followed by subscripts with the symbol ...

4

I simplified your model just a bit. sol = ParametricNDSolveValue[{ A1'[t] == -k1*A1[t] + k2*A2[t], A2'[t] == -k3*A2[t] + k4*A1[t] + k5*A3[t], A3'[t] == -k6*A3[t] + k7*A2[t], A1[0] == 0.23*10^4, A2[0] == 0.048*10^4, A3[0] == 0}, {A1, A2, A3}, {t, 0, 1}, {k1, k2, k3, k4, k5, k6, k7}] model[k1_, k2_, k3_, k4_, k5_, k6_, k7_, l_, m_, n_][t_] := ...

4

With uy0 defined in terms of x0 as Clear[uy0]; fuy0[x0_] := Solve[(H /. {x[t] -> x0, y[t] -> y0, ux[t] -> ux0, uy[t] -> uy0}) == H0, uy0][[1, 1, 2]] the criterion for a repeated orbit as f[xp_, tp_] := Module[{xx = x[xp, fuy0[xp]] /. solp, yy = y[xp, fuy0[xp]] /. solp, uxx = ux[xp, fuy0[xp]] /. solp, uyy = uy[xp, fuy0[xp]] /. solp}, ...

3

No dumb at all, Plot[{f1[x], f2[x]}, {x, -π, π}] As commented by @StephenLuttrell NSolve works with intervalls quite well; NSolve[f1[x] == f2[x] && -3 < x < 3, x] {{x -> -1.89549}, {x -> 1.89549}} As well FindInstance and FindRoot FindInstance[f1[x] == f2[x], x] // N // Chop {{x -> 1.89549}} FindRoot[f1[x] == f2[x], ...

3

Using recursion (NOTE that recursion can go either up or down) Clear[t] t[0] = 1; t[1] = 2; t[i_Integer?Positive] := t[i] = 2 t[i - 1] - t[i - 2]; t[i_Integer?Negative] := t[i] = 2 t[i + 1] - t[i + 2]; list1 = t /@ Range[-5, 5] (* {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6} *) The general solution is FindSequenceFunction[ Thread[Range[-5, 5] -> ...

2

For any given theta, you have a curve in phi/omega space. Thus, one way to visualize this is to plot in the phi/omega plane over all theta: sol = Solve[{-Cos[φ] Sin[2 θ] Sin[π Ω]^2 + Sin[θ] Sin[φ] Sin[2 π Ω], -2 Sin[θ] Sin[π Ω] (Cos[φ] Cos[π Ω] + Cos[θ] Sin[φ] Sin[π Ω]), -Cos[θ]^2 - Cos[2 π Ω] Sin[θ]^2} == {0, 1, 0}, {θ, φ, Ω}]; ...

2

Your code re-done: s = Solve[{ -x Cos[30 °] + y Cos[t ° ] == 0, x Sin[30 °] - y Sin[t ° ] == 50, -y Cos[t ° ] + z Cos[40 °] == 0, y Sin[t ° ] + z Sin[40 °] == 125}, {x, y, t, z}]; N.B. When entering a Degree token adjacent to a symbol, you must separate it from the symbol with a space. In the case of a numeric value, the space is ...

2

Solve, Reduce, NSolve can not solve this problem; this is a transcendental equation. It can only be solved numerically. \left\{\frac{e^{-2 \pi \sqrt{3-P}} \left(\sqrt{3-P}+i \sqrt{P}\right)-\sqrt{3-P}+i \sqrt{P}}{e^{-2 \pi \sqrt{3-P}} \left(\sqrt{3-P}+i \sqrt{P}\right)+\sqrt{3-P}-i \sqrt{P}}-\frac{i \left(1+e^{-2 i \pi \sqrt{P}}\right) ...

2

Edit Here is a solution that addresses energy level: (* parameter dep. system *) DE = DifferentialEquations[H, om, X, Y, UX, UY] ; (* function of initial coordinates for fixed end time *) f1[val_?NumberQ] := With[ {T=val}, ParametricNDSolveValue[ DE, {x[T],y[T],ux[T],uy[T]}, {t, 0, T}, {X,Y,UX,UY}, MaxSteps -> Infinity, ...

2

Interpolation can also be used and I present this just this somewhat 'over the top' to illustrate a number of ways to check, hone in initial values etc. I have 'simplified' the recursive function. fun[s_, x_] := x/(2 x - x^2 + (1 - x)^2/(s + 1)) r[s_, n_] := Nest[FullSimplify[fun[s, #]] &, 0.00001, n] tab = Table[{j, r[j, 50]}, {j, 0, 0.5, 0.001}]; if ...

1

For a beginner one hint: f[x_] = Exp[x] + Sin[x] - 4 // N; NestList[(# - f[#]/f'[#]) &, 1, 5] {1, 1.1351, 1.12999, 1.12998, 1.12998, 1.12998} Please read the documentation

1

You don't describe the problem you are having with the code you have, but I think I can guess. In Mathematica, functions like Sin use square brackets [] to delineate arguments. So your definition of your function f4 should be: f4[x_] := Piecewise[{{x Sin [(1/x)], -1 <= x < 0 || 0 < x <= 1}}, 0] You can then get a useful answer from NSolve: ...

1

It seems you don't need the other two equations, the second one by itself defines the curves: ContourPlot3D[ -2 Sin[θ] Sin[π Ω] (Cos[φ] Cos[π Ω] + Cos[θ] Sin[φ] Sin[π Ω]) == 0.99, {θ, 0, 2 π}, {φ, 0, 2 π}, {Ω, 0, 2}, PlotPoints -> 100, MaxRecursion -> 0] Another way to do it is to use Maxim Rytin's BoundaryStyle trick as described by Daniel ...

1

As march pointed out in the comment it would be better to use Array. Then you can define a function like f[i_] := Qcond[Kcond, x, T[i], T[i + 1]] == Qcond[Kcond, x, T[i + 1], T[i + 2]] Use Thread to write the conditional part 1 c1[n_] := And @@ Thread[Array[f, n, 0]] For the remaining part again use Thread c2[n_] := And @@ Thread[Array[T, ...

1

temp[Conc_, Kcond_, x_, h_] := NSolve[ And @@ Join[ {Qrad[Conc] == sigmaT0^4 + Qcond[Kcond, x, T[0], T[1]], Qcond[Kcond, x, T[19], T[20]] == sigmaT20^4 + Qconv[h, T[20]]} , Array[Qcond[Kcond, x, T[#], T[# + 1]] == Qcond[Kcond, x, T[# + 1], T[# + 2]] &, 19, 0] , T[#] > 0 & /@ Range[0, 20] ] , Table[T[k], {k, 0, 20}] ] There are ...

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