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How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc

You can also use Refine with Element : ...
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Logarithm of exponential

The assumption a > 0 is needed when Simplify is called: ...
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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 ...
• 64.5k
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Simplify with Assumptions Sqrt[(expr)^2]

It would appear that you can increase the number of assumptions variables that Mathematica will handle by altering a system option: ...
• 83.4k

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 ...
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Linearity Assumption

K1 /: Derivative[k_][K1] /; k >= 2 := (0 &)
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Defining the domain of positive real numbers

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 ...
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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 ...
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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 ...
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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 ...
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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: ...
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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 ...
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Why does Mathematica simplify $x/x\to1$?

Let's define f[x_] := Sqrt[x^2] FullSimplify[D[f[x],x], x ∈ Reals] (* ==> Sign[x] *) You can use f'[x] or ...
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Is there a way to simplify a symbolic expression at assignment, once-for-all?

You can use Evaluate to "force evaluation of the right-hand side of a delayed definition" (as stated in its documentation). For example ...
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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 ...
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$Assumptions = Element[{a, b, c, d, f}, PositiveReals] ; Simplify[Sign[a + b + c]] 1 ... • 350k 9 votes Accepted Behavior of Solve with$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 <...
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How can I use assumptions with FindRoot?

If you already know the interval on which you want to find one of your solution, you may use the instruction FindRoot[f[x]==0,{x,xmin,xmax}] Here, Mathematica ...
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Prove (or check) an expression is positive given constraints on variables?

Let's define: f[c_, d_, k_, toff_, ton_, V_] := "the expression equal to j" Instead of using toff, ...
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Why does Mathematica simplify $x/x\to1$?

fun = Sqrt[x^2]; dd = D[fun, x, x]; With V10 we can explicitly define: {dd, FunctionDomain[dd, x]} Or ...
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Simplifying conditional expressions using assumptions does not work

If you are willing to use Assuming, which acts by way of [$Assumptions](http://reference.wolfram.com/language/ref/$Assumptions....
• 264k
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Get Mathematica to Apply Chu-Vandermonde Convolution

Add to the the assumptions that 1 + a > b + d ...
• 121k