# Tag Info

Accepted

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 ...

### Logarithm of exponential

The assumption a > 0 is needed when Simplify is called: ...
• 20.4k
Accepted

### Using Assuming with Reduce

Assuming >> Details: Assuming affects the default assumptions for all functions that have an Assumptions option. Assumptions is not an option for ...
• 399k

### Assumptions allowing to calculate an elliptic integral

There are two different issues: suboptimal handling of elliptic integrals in Mathematica, this is why Integrate with appropriate assumptions doesn't provide ...
• 57.6k
Accepted

### Linearity Assumption

K1 /: Derivative[k_][K1] /; k >= 2 := (0 &)

### Complex number operations: telling Mathematica variables are real

The most complete and extendable answer is to define Conj[x_] := Refine[Conjugate[x], _Symbol ∈ Reals]; Then we get ...
• 1,525
Accepted

### Is there a way to automatically find the limits of integration given a set of constraints?

You can directly integrate over the region defined by the conditions: R = ImplicitRegion[x^2 + y^2 <= 1, {x, y}] Integrate[x^2, {x, y} ∈ R] π/4 If this ...
Accepted

### Assumption on the range of a function

EDIT: Simplified per suggestion from @BobHanlon. This constructs assumptions by finding all occurrences of pattern f[_] in the expression being simplified: ...
• 19.1k

### Using Assuming with Reduce

There is a warning in the docs for FullSimplify: Some of the transformations used by FullSimplify are only generically correct. ...
• 239k
Accepted

### Simplify is not simplifying a compound inequality as expected

Something I've noted about Mathematica, is it is best to be complete in defining variable domains. Also define i as an integer, which you might think it implicitly ...
• 7,183
Accepted

$Assumptions = Element[{a, b, c, d, f}, PositiveReals] ; Simplify[Sign[a + b + c]] 1 ... • 399k 10 votes Accepted ### Behavior of Solve with$Assumptions changed in 12.2

I think this is (seen by WRI as) an improvement, and the change is marked in the docs for Solve (noted in the comments). In the docs, it is also indicated how <...
• 239k

### Why Mathematica is treating the product of (specified) real variable as complex?

You are probably not using Mathematica correctly. First, you shouldn't put MatrixForm into SingularValueList. Second, the way ...
• 27.6k

### Simplify makes Mathematica forget that a matrix is Hermitian

There are two issues here. The first is that HermitianMatrixQ (in line with other Q functions) only applies quick checks that ...
• 16.8k

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• 160k
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• 399k
Accepted

### Get Mathematica to Apply Chu-Vandermonde Convolution

Add to the the assumptions that 1 + a > b + d ...
• 160k

### Specify range of variable in a equation

Add Assumptions : zw=FullSimplify[ Integrate[t1^2*E^((1 - h)*s0*t1), {t1, 0, T}], Assumptions -> 0 < h < 1 ] Addendum ...
• 54.9k
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 ...
• 239k
Accepted

### Mathematica tries to differentiate Abs[x] and this causes a problem?

Since Derivative[1][RealAbs][x] work, so we can use /. Abs -> RealAbs ...
• 77.8k
Accepted

### Making a substitution within a typeset integral

You could use the new in V 13.1 IntegrateChangeVariables int = Inactive[Integrate][Sqrt[Log[9 - x]]/(Sqrt[Log[9 - x]] + Sqrt[Log[x + 3]]), {x, 2, 4}] ...
• 147k
Accepted

### Why is Assuming[...] ignoring the assumption?

Assuming works by adding conditions to $Assumptions. Some functions make use of$Assumptions ...
• 3,570

### Defining the domain of positive real numbers

New in Mathematica 12 is PositiveReals (and others like NonNegativeIntegers, etc): ...
• 131k

### Forcing Mathematica's Integrate to give more general answers

If $\alpha$ is complex, then yes, The imaginary part of alpha should be strictly greater than zero. If $\alpha$ is real, does it converge? If you complete the contour in the complex plane, with a ...
• 2,166

### Taking real and imaginary parts after reciprocal

If x=0, then 1/0 is ComplexInfinity. If you add the assumption that ...
• 1,309
Accepted

### Assuming a variable is imaginary

This needs FullSimplify. FullSimplify[x + Conjugate[x], Re[x] == 0] 0
• 108k