18
Are you looking for Subtract?
eq=x>=y
Subtract@@eq>=0
gives:
x-y>=0
Edit
If one wants a function, which keeps the order sign and adds the 0, one may use:
oneSide=(Head[#][Subtract@@#,0]&)
and call e.g. eq//oneSide
17
The proof of the original statement that $f(x)\equiv x\sin\frac{\pi}{x}$ is a monotonically increasing function of $x$ for $x>1$ can be done as follows:
First, we show that the second derivative $f''(x)$ of the function is negative:
Simplify[D[x Sin[π/x], x, x] < 0, Assumptions -> x > 1]
True
This means that the first derivative $f'(x)$ is ...
15
Mathematica often responds well when provided a little expert assistance. Let's focus on techniques that have a wide application rather than just to this problem.
Can the function be decomposed into simpler pieces? Yes, obviously: $f(x)$ is the product of $x$ and $\sin{\pi / x}$. Both are obviously increasing for $x \in [1,2]$. After that, $\sin{\pi / x}...
13
From the documentation:
Less gives True or False when its arguments are real numbers.
Regardless of what x or y are, it is clear that {x,y} is not a real number, so Less doesn't do anything with it (for the most part, some simplifications may still be performed). Note that other functions may also treat {x,y}<0 specifically, and that you can ...
13
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 cases where Mathematica will accept a list with <, ==, >, such as
Solve[{x, y} < 2 && y == x^2]
(* {{y -> ConditionalExpression[x^2, -Sqrt[2] < x < Sqrt[2]]}} *)
This ...
12
This approach works by using the fact that an inequality or equality can be traversed by Map in the same way that a regular list can. It can take an arbitrary inequality or equation eqn, and you don't have to know in advance whether it's >, < or anything else. First I define the equation eqn, and then I use the fact that the second part of eqn is the ...
12
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 I feel you mean assumption), I say it's an expression, a Boolean expression. An expression like x>0 doesn't state that x is greater than 0. It's an undetermined claim about x that is True when x is indeed greater than 0, and ...
answered Jan 11 '17 at 19:21
Sjoerd C. de Vries
62.5k12 gold badges170 silver badges308 bronze badges
11
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 inequalities in several variables:
f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]
2.625
Edit:
This one-argument version treats all symbols in the first argument as variables:
f[ineq_] := f[ineq, ...
10
SetSystemOptions[
"ReduceOptions" -> {"DiscreteSolutionBound" -> 10000000}];
f[n_] := Module[{eqn},
eqn = (n1 + n3 + n5 == n2 + n4 + n6) && (n1 >=
n2) && (n1 - n2 + n3 - n4 >= 0) && (n1 <= n) && (n1 - n2 + n3 <=
n) && (n1 - n2 + n3 - n4 + n5 <= n) && (2 n <=
n1 + ...
10
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 described when we fix one of the two variables, therefore we can define e.g.:
r[t_] := Reduce[ eq[t, x] < 1/100 && t > 0 && x > 0, {x}]
We need the function r because in ...
9
Since Mathematica 11.3 you can use SubtractSides, that works for equations and inequalities, for example
eqns = {
x >= y,
x^2 == b c + c,
2 x < x + 1
};
SubtractSides /@ eqns
(* {x - y >= 0, -c - b c + x^2 == 0, -1 + x < 0} *)
9
You can do :
p = ImplicitRegion[y <= 3/10 x + 18 && y > x^2/8, {x, y}]
points = Reduce[Element[{x, y}, p], {x, y}, Integers]
pp = Cases[points, x == xx_ && y == yy_ -> {xx, yy}]
pp // Length
(* 286 *)
Show[RegionPlot[p], ListPlot[pp]]
answered Aug 22 '14 at 12:50
b.gates.you.know.what
18.5k2 gold badges35 silver badges72 bronze badges
9
con = Thread[{a, b, c} > 1/2 + Sqrt[5/4]];
Simplify[Reduce[Flatten[
{a b c (a + b + c) > 3 a b c + a b + b c + c a, con}],
{a, b, c}], con]
True
Graphically,
Show[
RegionPlot3D[
a b c (a + b + c) > 3 a b c + a b + b c + c a,
{a, 0, 3}, {b, 0, 3}, {c, 0, 3},
PlotStyle -> Directive[Blue, Opacity[0.25]]],
RegionPlot3D[And @@ con,
...
9
As I stated in my comment below the post, the filling syntax used by the OP is correct. The behavior seen in the plot is a bug.
A workaround is to simply increase the number of plot points. The following works:
Plot[{x^2, x^4}, {x, -2, 2},
PlotRange -> {-1, 5}, PlotStyle -> {Automatic, Red},
Filling -> {1 -> {{2}, {LightBlue, White}}}...
9
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 attempt to decide whether the system has solutions. (Simplify proves that an inequality follows from the assumptions by showing that assumptions && Not[...
8
eqn = y <= 3/10 x + 18 && y > x^2/8;
sol = Reduce[eqn, {x, y}, Integers];
Length @ sol
(* 286 *)
points = {x, y} /. {ToRules[sol]}; (* thanks: BobHanlon *)
RegionPlot[eqn, {x, -15, 18}, {y, -5, 25},
GridLines -> {Range[-15, 18], Range[-5, 25]},
PlotStyle -> Directive[{Opacity[0.5], Red}],
Epilog -> {...
8
Since Reduce doesn't seem to like the inequality, I tried FullSimplify with Assumptions instead. This works in three steps:
differenceByTerm =
SeriesCoefficient[(1 + x)^n - (1 + n x), {x, 0, m}]
$$
\cases{
0 & m=0 \\
\binom{n}{m} & m>1 \\
0 & \text{True} \\
}$$
FullSimplify[
differenceByTerm >= 0,
Assumptions -> n > 1 ...
8
And @@ Thread[{a, b} > {c, d}]
(* a > c && b > d *)
8
re = Reduce[f1[x, y] < f2[x, y], {x, y}, Reals]
Reduce[f1[x, y] > f2[x, y], {x, y}, Reals]
x < 1 && -Sqrt[1 - x] < y < Sqrt[1 - x]
(x <= 1 && (y < -Sqrt[1 - x] || y > Sqrt[1 - x])) || x > 1
Edit: You can also directly plot the reduced result.
RegionPlot[re, {x, -10, 5}, {y, -5, 5}, ImageSize -> Small]
8
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 + r[3, 5]^-α +
r[4, 5]^-α
Alternatively,
Sum[Boole[1 <= j < k <= n]/r[j, k]^α, {j, 1, n}, {k, 1, n}]
r[1, 2]^-α + r[1, 3]^-α + r[1, 4]^-α +
r[1, 5]^-α + r[2, 3]^-α + r[2, 4]^-α ...
7
Here is another solution using V10 functionalities:
region = ImplicitRegion[y <= 3/10 x + 18 && y > x^2/8, {{x, -15, 18}, {y, -5, 25}}];
lis = Tuples[{Range[-15, 18], Range[-5, 25]}];
We create a RegionMemberFunction
rm = RegionMember[region];
Now we select from lis the points that are in the region:
in = Select[lis, rm];
Length @ in
...
7
Basically it amounts to using CoefficientArrays along with a bit of massaging for sense of inequality.
getLinearForm[ineqs : {(_LessEqual | _GreaterEqual) ..}, vars_] :=
Module[
{mults, lpolys, rhs, lhs},
mults = Map[Head, ineqs] /. {LessEqual -> 1, GreaterEqual -> -1};
lpolys = mults*Apply[Subtract, ineqs, {1}];
{rhs, lhs} = Normal[...
7
Yes, Mathematica can prove these inequalities symbolically. To be more precise, it can Reduce them to True. Generating human-readable proofs is also possible but that's one broad topic.
First, we'll use a “coordinate change”. Notice that your expressions for R, s, r are all symmetric in x, y, z. This hints that we might benefit from using symmetric ...
7
Yes, it is documented that the result of Reduce[expr, vars] always describes exactly the same mathematical set as expr, i.e. the result of the reduction is equivalent to the original system.
Another way to state the above is
In[1]:= s1 = c > 0 && ((d <= -2 - c && a > 0 && b > (-c - d)/a) ||
(-2 - c < d ...
7
Posting this here as community wiki because we answers should not stay only as comments.
You have many options
Using Resolve and ForAll
Resolve[ForAll[x, x > 0, Abs[x] == x]]
Using Refine
Assuming[x > 0, TrueQ[Refine[Abs[x] == x]]]
But assumptions can be placed inside Refine, as in
Refine[Abs[x] == x, x > 0]
Using Simplify
Simplify[Abs[x] =...
7
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 boundary?
ContourPlot[{f1[x, y] == f2[x, y], x == 1 - y^2}, {x, -2, 2}, {y, -2, 2},
ContourStyle -> {, Dashed}]
Indeed
eq = f1[x, y]^2 == f2[x, y]^2 // FullSimplify
eq /. x -> 1 - y^2 // FullSimplify // PowerExpand
(...
7
If you are OK with turning the Ors into Columns, you can do something like:
result = Reduce[
h1 >= 0 && h2 >= 0 && 2*x >= 0 && -m + h1 + y >= 0 &&
m + x - y >= 0 && h2 - x + 2 y >= 0, {x, y}];
TraditionalForm[
result //.
{Or -> (Column[#, Right, Background -> {{White, LightGray}}, ...
6
I think that this problem (like most algebraic manipulations) is best approached with pattern matching and rule replacement than by directly exploiting the index/list structure that inequalities share with practically every Mathematica expression. Both approaches will work, but rule replacement is, in my opinion, clearer, more precise, more flexible, and ...
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