Tag Info

New answers tagged

5

You can gain insight by evaluating and then simplifying the indefinite integrals. Integrate[(1 + b x + c y)/(1 + e x + f y + I η), y, x, Assumptions -> {x ∈ Reals, y ∈ Reals, b ∈ Reals, c ∈ Reals, e ∈ Reals,f ∈ Reals, η ∈ Reals, η != 0}]; ans = Collect[%, {ArcTan[η/(1 + e x + f y)], Log[1 + e^2 x^2 + 2 f y + f^2 y^2 + 2 e (x + f x y) + η^2]}, ...


5

One way to approach this is to look for the source of the problem. In this case, the outer integral (on y) is irrelevant to the problem, because even the simpler integral int = Integrate[(1 + b x)/(1 + e x + I η), {x, -1/2, 1/2}, Assumptions -> {b ∈ Reals, c ∈ Reals, e ∈ Reals, η ∈ Reals, η != 0}] gives a conditional expression. But now the answer ...


2

There's a misunderstanding here. The third "dom" argument is not simply a set over which we solve the equation. There are only a few choices that can be used for the domain argument, and they have very specific effects on how Solve works. An example from the documentation: If dom is Reals, or a subset such as Integers or Rationals, then all constants ...


1

{Solve [ x^2 == 1, {x}], Solve [ x^2 == 1 && x > 0, {x}]} (* {{{x -> -1}, {x -> 1}}, {{x -> 1}}}*)


1

potentialOfSphere[x_] = Integrate[ -m/(4/3 Pi r^3)*g/Sqrt[(x - r1 Cos[ϕ])^2 + (r1 Sin[ϕ])^2]*r1^2 Sin[ϕ], {r1, 0, r}, {θ, 0, 2 Pi}, {ϕ, 0, Pi}, Assumptions -> {x > 0, r > 0}] (gm(-2*r^3 + (r - x)^2*(2*r + x)* HeavisideTheta[r - x]))/ (2*r^3*x) For x > r, this reduces to potentialOfSphere[x] // Simplify[#, x > ...


3

It appears Surd is not integrated into Quantity (and it probably should be). This auto simplifies: r = Quantity[10, "Meters"]; v = 4/3 r^3 Pi; (%*3/4/Pi)^(1/3) Quantity[10, "Meters"]


0

The problem seems like a bug, a rather silly one on the surface. Consider the returned value in the OP's first bit of code. When it is simplified using the equivalent assumption as coded by the OP, it does not return True. A simple transformation of the inequality gets us to the result. Simplify[(-1)^L Sqrt[1 - a^2] Cos[2 b] <= 1, 1 - (-1)^L Sqrt[1 ...


2

You can add ComplexExpand to the transformations FullSimplify will try (since a is real). But FullSimplify seeks to minimize the complexity of the expression, and the starting expression is a local minimum, To get it over the hump, we can penalize Abs. Assuming[0 <= a <= π/2, FullSimplify[Abs[Sin[4 a]] == 2 Abs[Cos[2 a]] Sin[2 a], ...


3

As I was recently informed by Wolfram support, Simplify is not guaranteed to fully solve equalities or inequalities. Take this example: FullSimplify[Log[x] > 1, Element[x, Reals]] (* Log[x] > 1 *) Reduce[Log[x] > 1, x, Reals] (* x > E *) We can also get your equation to simplify by using Reduce. Reduce[Abs[Sin[4 a]] == 2 Abs[Cos[2 a]] Sin[2 ...


4

You can use Evaluate to "force evaluation of the right-hand side of a delayed definition" (as stated in its documentation). For example f[x_] := Simplify[Sqrt[x^2], x > 0] Definition@f f[x_] := Simplify[Sqrt[x^2], x > 0] g[x_] := Evaluate@Simplify[Sqrt[x^2], x > 0] Definition@g g[x_] := x


0

The following does not answer the Question, but may shed some light on it. For convenience, name the two equations eq1 and eq2, respectively. eq1 = Abs[1 - (-1)^L Sqrt[1 - a^2] Cos[2 b]] == 1 - (-1)^L Sqrt[1 - a^2] Cos[2 b]; eq2 = Abs[1 - E^(I π L) Sqrt[1 - a^2] Cos[2 b]] == 1 - E^(I π L) Sqrt[1 - a^2] Cos[2 b]; The essence of the matter appears to be ...



Top 50 recent answers are included