New answers tagged

1

One idea is to give a derivative definition: x /: Derivative[n_?(GreaterThan[1])][x] := b x'[#] - c & Then: a b x'[t]^2-a c x''[t] a b x'[t]^2 - a c (-c + b x'[t]) and: x'[t] x''[t] - b x'[t] - c == 0 -c - b x'[t] + x'[t] (-c + b x'[t]) == 0


2

Clear["Global`*"] eqs1 = a b x'[t]^2 - a c x''[t]; eqs2 = x''[t] - b x'[t] - c == 0; eqs2 can be solved for x[t] using DSolve sol = DSolve[eqs2, x, t][[1]] (* {x -> Function[{t}, -((c t)/b) + (E^(b t) C[1])/b + C[2]]} *) Verifying, eqs2 /. sol // Simplify (* True *) Substituting into eqs1 expr = eqs1 /. sol // Simplify (* a b (-c E^(b t) ...


2

the output should like $eq1 = ab\dot{x}^2 -ac(b\dot{x}+c)$ I am sure there many ways to do this in Mathematica. One way could be Clear["Global`*"] eqs1 = a*b*x'[t]^2 - a*c*x''[t] eqs2 = x''[t] - b*x'[t] - c == 0 eqs1 /. First@Solve[eqs2, x''[t]] If you do not use First@ above, you'll get same answer but in a list eqs1 /. Solve[eqs2, x''[t]] ...


0

If you have a single relationship that you would like to enforce tp[v] ** m ** k ** v -> tp[v] ** k ** m ** v then you can apply that as a rule and your expression will simplify. For example: expr = tp[v] ** m ** k ** v - tp[v] ** k ** m ** v NCReplaceAll[expr, tp[v] ** k ** m ** v -> tp[v] ** m ** k ** v] will evaluate to zero. However, I suspect ...


0

In NCAlgebra NCDot[{a, b}, {c, d}] evaluates to a ** c + b ** d as you want. NCDot is the familiar matrix product that takes into account noncommutative entries. Another example: NCDot[{{a, b}, {c, d}}, {e, f}] evaluates to: {a ** e + b ** f, c ** e + d ** f}


3

In calculus class we might differentiate both sides: ApplySides[D[#, x] &, Integrate[a[x], x] + Integrate[b[x], x] == Integrate[c[x], x] ] (* a[x] + b[x] == c[x] *) (Avoid beginning your symbols with a capital. For instance C is a Protected symbol in Mathematica.)


6

Clear["Global`*"] Solve for Bessel functions in terms of Hankel functions sol = Solve[(# == FunctionExpand[#]) & /@ {HankelH1[n, z], HankelH2[n, z]}, {BesselJ[n, z], BesselY[n, z]}] (* {{BesselJ[n, z] -> 1/2 (HankelH1[n, z] + HankelH2[n, z]), BesselY[n, z] -> -(1/2) I (HankelH1[n, z] - HankelH2[n, z])}} *) Convert solutions to ...


7

For a simple expression that doesn't require simplification, it should be enough to use ReplaceAll with a suitable substitution. B2H = ReplaceAll[ { BesselJ[a_, b_] :> (HankelH1[a, b] + HankelH2[a, b])/2, BesselY[a_, b_] :> I*(HankelH2[a, b] - HankelH1[a, b])/2 } ] BesselJ[\[ImaginaryI]*m, k*r] //B2H (*(HankelH1[\[ImaginaryI] ...


4

This question arises from time to time on this site. Unfortunately, I have no references to the previous discussions. In general, Mma has its internal order to write the expressions. The use of the TraditionalForm may partially change this order, but not completely. It is impossible to violate this order without introducing some limitations to the expression....


0

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


1

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 && b == -2 c + k && a == 0) || (d == e && c == f && b == -2 f + k)*)


1

One possible way Clear["Global`*"] eq1 = (a + b + c) x^2 + (a c + b) x + d == 0; eq2 = (a + k) x^2 + (a f + b) x + e == c x^2; eq = First@eq1 - Last@eq1 == First@eq2 - Last@eq2; (Thread@ApplySides[CoefficientList[#, x] &, eq]) // Column Or Clear["Global`*"] eq1 = (a + b + c) x^2 + (a c + b) x + d == 0; eq2 = (a + k) x^2 + (a f + b) x ...


2

Setup: b1 = {{0, 0, 0}, {0, 0, 1}, {0, -1, 0}}; b2 = {{0, 0, -1}, {0, 0, 0}, {1, 0, 0}}; b3 = {{0, 1, 0}, {-1, 0, 0}, {0, 0, 0}}; sys = {xt1, xt2, xt3} == {x1, x2, x3} . MatrixExp[t1 b1] . MatrixExp[t2 b2] . MatrixExp[t3 b3]; Eliminating three variables from the three equations sequentially works quickly: Quiet[ Eliminate[ Eliminate[ Eliminate[...


2

The quite fastest way is to use Solve, which is not sensitive if some variables are not independent of each other, add identities for Sin, Cos and eliminate all Sin und Cos at the same time. Solve[{xt1 == Cos[t3] (x1 Cos[t2] + (x3 Cos[t1] + x2 Sin[t1]) Sin[t2]) - (x2 Cos[ t1] - x3 Sin[t1]) Sin[t3], xt2 == Cos[ t3] (x2 Cos[t1] - x3 Sin[t1]) + (x1 ...


3

You could always make a Rule and choose to apply it or not. cRule = {c -> Sqrt[a^2 - b^2]}; F[x_, y_, z_] = c*x + c^2*y + c^3*z (* c^3 z + c^2 y + c x *) or F[x, y, z] /. cRule (* x Sqrt[a^2 - b^2] + y (a^2 - b^2) + z (a^2 - b^2)^(3/2) *)


0

If you literally wish to use {e == a/d, f == b/d, g == c/d}, this also works. Expand[#/d & /@ (a*x + b*y + c*z == d)] /. (Reverse /@ {e == a/d, f == b/d, g == c/d} /. Equal -> Rule) (* e x + f y + g z == 1 *)


2

Clear["Global`*"] eqn = a*x + b*y + c*z == d; Using Eliminate Assuming[{x != 0, y != 0, z != 0}, Eliminate[{eqn, e == a/d, f == b/d, g == c/d}, {a, b, c, d}] // Simplify] (* e x + f y + g z == 1 *)


0

Consider using Simplify and Assumptions Simplify[Conjugate[a + I*b], Assumptions -> {{a, b} \[Element] Reals}] Result a - I b


0

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*)


1

eq = a*x + b*y + c*z == d Simplify[eq /. {a -> e d, b -> f d, c -> g d}, d != 0] (* e x + f y + g z == 1 *)


9

equation = y + z == D[Integrate[1/(x^3 + 1), x], x] You can use MapAt: MapAt[Simplify, equation, {2}]


4

I'd recommend going as far as possible with analytic calculations and only using numerics at the end. Especially numerical derivatives are notoriously difficult to calculate exactly. Define the function $A$ without numerics: A[u_, x_, t_] = AiryAi[x - u*t + I*u*t - t^2] * Exp[I*t (x - u*t - t^2 + u/2)] * Exp[x - u*t - t^2]*Exp[I*(x - u*t)]; Define its ...


2

Edit ... (Contrary to first result, true without conditions) Splitting b and c into real and imaginary part together with Reduce helps to show second equation beeing True. red2 = Reduce[ b1 \[Element] Reals && b2 \[Element] Reals && c1 \[Element] Reals && c2 \[Element] Reals && Cos[Sqrt[-(b + c)]] == Cosh[Sqrt[(b + c)]] /....


2

Also Simplify can do the job, if you feed appropriate TransformationFunctions {g1[-Subscript[y, 1] - Subscript[y, 2] - Subscript[y, 3] - Subscript[y, 4]] := -r1, g2[Subscript[y, 1] + Subscript[y, 2] + Subscript[y, 3] + Subscript[y, 4]] := r1, g3[Subscript[y, 1]^2 + Subscript[y, 2]^2 + Subscript[y, 3]^2 + Subscript[y, 4]^2] := r2}; Simplify[result, ...


3

$i_2 = \frac{b_3}{p_1} j$ follows directly from setting ${dj \over dt}=0$. You could do that manually, or in EcoEvo, give SolveEcoEq a list of variables to solve for and use QSS -> True (otherwise non-specified variables will be assumed equal to zero). SolveEcoEq[{j}, QSS -> True] (* {{j -> (i2 p1)/b3}} *) You might also try: SolveEcoEq[{a}, QSS -&...


5

eqPts is the equilibrium points in your question. Start by using Simplify LeafCount /@ {eqPts, eqPts2 = eqPts // Simplify} (* {1958, 448} *) The variables are vars = Variables[Level[eqPts2, {-1}]] (* {a, b1, b2, b3, b4, i1, i2, j, p, p1, p2, q, s, β1, β2, \ ϵ, μ, ν, ξ1, ξ2} *) Convert the replacement rules in eqPts2 into the corresponding equations eqns =...


1

Before feeding a function to "Simplify" it will be evaluated. Now, consider: Clear[Q]; Q[x_] := Sign[x] /; x > 1 Q[x] (*Q[x]*) Clear[Q] Q[x_] := Piecewise[{{Sign[x], x > 1}}] Q[x] As you see, the difference is that in the first stage of evaluation Q[x] is not replaced in the first case and replace in the second case. If you ensure that Q[x] ...


3

Get an overview te = (w == wP || z < x) && (w == wP || w > x + 2 z) && (wP == 4 + 2 x + z || z >= x || w <= x + 2 z) && z == zP && (x + z == yP || z < x) && 2 x + z == xP && (x + 2 z == yP || z >= x); TraditionalForm[ te //. Or -> Composition[(Column[#, Right, Background -> {{White,...


1

First note, that $Pre without an attribute "HoldFirst" or "HoldAll" is the same as $Post. Therefore,if you input: "Sign[a Eps] /. a -> 3". This will evaluate to Sign[3 Eps]. And this evaluates to Sign[Eps] because 3 is positive. Finally, what $Pre sees is therefore: Sign[Eps]. On the other hand, an input "Sign[a Eps]&...


2

The problem here is that EvenQ@ximmediately evaluates to False. As you note, if you replace this by Mod[x,2]==0 then PiecewiseExpand works fine. In my view, what you really need here is a domain, say adding Evens and Odds—just as NonNegativeIntegers and friends were added in 12.0—so that you can write Element[x,Evens] instead. As a related example, instead ...


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