13 votes

Is it possible to have Mathematica move all terms to one side of an inequality?

Since Mathematica 11.3 you can use SubtractSides, that works for equations and inequalities, for example ...
rhermans's user avatar
  • 36.5k
13 votes
Accepted

Reduce not making full use of list of assumptions?

What I believe is the issue here is what a statement in your Reduce like c ∈ Reals actually means. You say it's a condition (and ...
Sjoerd C. de Vries's user avatar
13 votes

What does `{x,y}<0` mean?

Generally, this is an invalid use of <. Both the left and the right-hand side should be expressions representing real numbers. However, there are a few special ...
Szabolcs's user avatar
  • 235k
12 votes
Accepted

What does `{x,y}<0` mean?

From the documentation: Less gives True or False when its arguments are real numbers. ...
eyorble's user avatar
  • 9,458
12 votes

Calculating the width of the interval defined by an inequality

f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var], Length[Flatten[{var}]]] f[1 <= x <= 2.5, x] 1.5 This works also for some systems of ...
Henrik Schumacher's user avatar
12 votes
Accepted

Calculating union of two sets of reals

You can try Reduce[{-(1/4) < x <= 0 || 0 < x <= 1/2 || x > 1/2}, x] x > -(1/4) ...
Laurenso's user avatar
  • 1,032
11 votes

Checking inequalities: How can $x>0,y>0$ yet $x+y$ indeterminate?

The assumptions mechanism used by Simplify has a bound on the number of variables in a system of nonlinear inequalities. If the number of variables exceeds the bound, the assumption mechanism does not ...
Adam Strzebonski's user avatar
10 votes
Accepted

Can Mathematica solve $|-x\cdot\exp(-0.5x^2/b^2)/b^2|<0.01$?

If we want exact values we should work with them from scratch. eq[t_, x_] := Abs[-x Exp[-1/2 x^2/(t^2)]/(t^2)] The boundary of the set of our interest can be ...
Artes's user avatar
  • 57.3k
9 votes
Accepted

Make Reduce produce nicer output

If you are OK with turning the Ors into Columns, you can do something like: ...
chuy's user avatar
  • 11.2k
9 votes

How can I test if two inequalities have a common part?

Does a value of x which fulfils both conditions exist? ...
kirma's user avatar
  • 19.1k
8 votes
Accepted

Relational operators on vectors

And @@ Thread[{a, b} > {c, d}] (* a > c && b > d *)
Dr. belisarius's user avatar
8 votes
Accepted

Summation with inequalities

n = 5; Sum[1/r[j, k]^α, {k, 1, n}, {j, 1, k - 1}] r[1, 2]^-α + r[1, 3]^-α + r[1, 4]^-α + r[1, 5]^-α + r[2, 3]^-α + r[2, 4]^-α + r[2, 5]^-α + r[3, 4]^-α + ...
kglr's user avatar
  • 395k
8 votes
Accepted

Solve for parameters so that a relation is always satisfied

What do you prefer: ...
user64494's user avatar
  • 26.4k
8 votes
Accepted

Why isn't this expression returning true to being positive when it is clearly positive?

Simplify is at root an expression tree minimizer, equipped with some algebraic and logical transformations. As such, it may have the transformations needed to reach ...
Michael E2's user avatar
  • 236k
8 votes
Accepted

Why does MMA fail to find a solution but WA does?

First, my stab at W|A gave a different output than the OP (something like what I show for Reduce below, in fact). Second, I'm not sure how to read the W|A output ...
Michael E2's user avatar
  • 236k
8 votes
Accepted

How do I adjust the form of the inequality?

{barray, marray}=CoefficientArrays[Equal@@@Flatten@ieq,y]; marray.{y}>=-barray//Thread
xzczd's user avatar
  • 66.2k
7 votes
Accepted

Strange behavior with Inequality and pattern matching

The issue is pre-evaluation of the pattern. For the ones that evaluate to False: Inequality[___] (* True *) and ...
march's user avatar
  • 23.5k
7 votes

Restrictions on variables of two functions in order to be larger and smaller relative to each other

re = Reduce[f1[x, y] < f2[x, y], {x, y}, Reals] Reduce[f1[x, y] > f2[x, y], {x, y}, Reals] ...
ciao's user avatar
  • 25.8k
7 votes

Volume of a 3D implicit region

I think the main issue is the axes bounds are quite disproportionate and that's effecting the sampling. Here's your region scaled to the unit cube: ...
Greg Hurst's user avatar
7 votes
Accepted

Isolate variable in inequality

Try Reduce[Abs[2 - x] > 2, x, Reals] (*x < 0 || x > 4*)
Ulrich Neumann's user avatar
7 votes

Can't this inequality really not be solved with Reduce?

We can change variable. f[n_, a_] = n + a*ProductLog[-E^(-n/a)]; f[n, a] <= 1 /. n -> t*a // Simplify ...
cvgmt's user avatar
  • 72.9k
7 votes

How can I test if two inequalities have a common part?

...
Syed's user avatar
  • 54k
6 votes
Accepted

Reduce a complex inequality

First of all you shouldn't (if possible) use approximate numbers working with symbolic functionality like a very sophisticated function Reduce. Before seeking the ...
Artes's user avatar
  • 57.3k
6 votes
Accepted

Problems with integrating over a region

You need to be consistent with the order of the variables defining the region: ...
b.gates.you.know.what's user avatar
6 votes
Accepted

Ask Mathematica if there exists a value that satisfies conditions

Mainly just to see how Reduce@Exists[..] stacks up against FindInstance[]. I suspect the heuristics of ...
Michael E2's user avatar
  • 236k
6 votes

Strange behavior with Inequality and pattern matching

Inactivate is helpful in analyzing this problem. ...
m_goldberg's user avatar
  • 108k
6 votes
Accepted

Restrictions on variables of two functions in order to be larger and smaller relative to each other

By trial and error, if we plot Plot3D[{f1[x, y], f2[x, y]}, {x, -2, 2}, {y, -2, 2}] we see that x== 1-y^2 is the ...
chris's user avatar
  • 23k
6 votes
Accepted

How to solve $n\left(\frac{n}{n+1}\right)^n>1$ for $n\in\Bbb N$?

Reduce works like a charm: Reduce[x*(x/(1 + x))^x > 1 && x > 0, x, Integers] x ∈ Integers && x >= 3
Julien Kluge's user avatar
  • 5,335
6 votes

Get the inequalities from answers

First of all I get five solutions for your system from Solve: ...
Mr.Wizard's user avatar
  • 272k
6 votes
Accepted

Plot surface defined by inequality

For arbitrary implicite regions it is easy to use: ...
Vitaliy Kaurov's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible