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Questions about the function Assuming[], the $Assumptions system variable, and the Assumptions option of functions such as Simplify[] and Integrate[].
2
votes
Accepted
How to make Maximize function successfully maximize in general?
Although, as the OP noted,
Maximize[{Log[x] + Log[y], x + y == 10 && 0 < x && 0 < y}, {x, y}, Reals]
returns unevaluated,
Maximize[{Log[x] + Log[y], x + y == 10 && 0. < x && 0. < y}, {x, y}, Reals …
2
votes
Accepted
Failing to give assumptions to find the solution of a Ricatti equations
Although Jim Baldwin's solution is entirely satisfactory (+1), it is possible to obtain the desired expression without the additional assumptions. … Flatten@DSolve[{a (FC[h])^2 - 2 b FC[h] + c - D[FC[h], h] == 0},
FC[h], h, Assumptions -> b^2 - a c > 0]
(* (b - Sqrt[b^2 - a c] Tanh[Sqrt[b^2 - a c] h + Sqrt[b^2 - a c] C[1]])/a *)
Now, obtainingC …
1
vote
VectorFieldPlot Assumptions - restricted domain
Use VectorPlot instead.
VectorPlot[{x/Sqrt[x^2 + y^2], y/Sqrt[x^2 + y^2]}, {x, -5, 5}, {y, -5, 5}]
2
votes
Mathematica does not distribute conjugation when used with Assuming sometimes?
Like the OP, I find that
Assuming[x_ ∈ Reals, FullSimplify[Conjugate[Exp[I x1] x2]]]
(* E^(-I x1) x2 *)
works, but
Assuming[x_ ∈ Reals, FullSimplify[Conjugate[Exp[I x1] x2 + x3]]]
(* E^(I x1) x2 …
0
votes
Simplifying expressions with Abs
Consider first the effect of making the first three assumptions listed in the Question
FullSimplify[{eq1[[1]], eq1[[2]], eq1}, -1 <= a <= 1 && 0 <= b <= π/4 && L ∈ Integers]
FullSimplify[{eq2[[1]], eq2 …
3
votes
Integral evaluation error
Problems of this sort are posted from time to time in Mathematica SE. Multiple instances of Nintegrate are nested one inside another, and an inner integrand contains one of the outer variables of int …
7
votes
evaluation of the sum of KroneckerDelta
the assumptions entirely also works, although not so cleanly. … But,
$Assumptions = 0 <= m <= nn && m ∈ Integers;
FullSimplify[Sum[KroneckerDelta[m, n] f[n], {n, -Infinity, nn}]]
does not. This may be a bug. …
1
vote
Integration result with incomplete beta function
As I commented above, there is nothing wrong with the result obtained in the Question. To make this point more clearly, consider a specific case,
ans /. {N1 -> 2, N2 -> 2, dt1/(dt2 k) -> x}
(* -30 B …
6
votes
Accepted
Well-defined symbolic integral leading to ConditionalExpression
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 ! …
5
votes
RSolve with assumptions
Workaround
A workaround based on a suggestion provided by Wolfram Technical Support for this particular problem is
SetOptions[PowerExpand, Assumptions -> 0 < a < 1];
FullSimplify[k[t] /. … Incidentally, SetOptions[PowerExpand, Assumptions -> True]; also seems to work. …
2
votes
Trouble evaluating this definite integral, either directly or with indefinite + limits
With it fixed and R > 1 chosen, consistent with the code in the question, the following returns an answer in just a few minutes for Version 12.1.1
Integrate[x*sigx, x, Assumptions -> 0 < x < 1 && R > 1 … R -> 2, s[0] == 0}, s[x], {x, 0, 1}];
Incidentally, the corresponding definite integral,
Integrate[x*sigx, {x, 0, y}, Assumptions -> 0 < y < 1 && R > 1]
returns unevaluated after several minutes. …