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}, Simplify @ Conjugate[e]], DownValues[Conjugate] = {} ]; Protect[Conjugate]; ) Then, turn on conjugate simplification: SetConjugateSimplification[True] Now, Conjugate will use assumptions … specified with $Assumptions or Assuming: $Assumptions = c10 ∈ Reals; Conjugate[c10] c10 Restore the usual Conjugate behavior: SetConjugateSimplification[False] Another possibility is to use …

One possibility is to just replace your variable with one whose powers are all explicit integers, and then use Series: e = PowerExpand[ a^(-c2)/((a*c1)^(c2) + c3) /. a -> z^(1/c2), Assumptions …

I think using Reduce would be a better approach, although it's not completely straightforward: Reduce[x^a < 1 && a > 0 && x > 1, Reals] False (updated with another approach using Resolve) Ano …

You could try using DSolveValue. First, define your sum using inactive integrals: h[α_] := Inactive[Integrate][x g'[x],{x,0,α}] + Inactive[Integrate][α g'[x],{x,α,∞}] Even though the integrals are …

Solve doesn't make use of Assumptions. Also, your first assumption doesn't mean anything. …

You need to use Update to tell Mathematica that hidden changes have been made that can affect results: THM=TensorDimensions[M]; Assuming[Element[M,Vectors[n]],THM] TensorDimensions[M] Using Up …

On the other hand, you can use explicit assumptions for x1 and x2 only: PowerExpand[Sqrt[x1^2] + Sqrt[x2^2] + Sqrt[x3^2], Assumptions->x1<0&&x2>0] -x1 + x2 + E^(I π Floor[1/2 - Arg[x3]/π]) x3 In … On the other hand, if you include a non-default Assumptions option, then the output of PowerExpand will be correct (for the given assumptions). …

ToRadicals supports the undocumented option Assumptions. … Making use of this option gives: sol = Solve[x^3==a^2,x,Reals] ToRadicals[sol, Assumptions -> a>0] {{x -> Root[-a^2 + #1^3 &, 1]}} {{x -> a^(2/3)}} …

Per the documentation you can use Esc ints Esc to enter the StandardForm symbol for Integers

New in Mathematica 12 is PositiveReals (and others like NonNegativeIntegers, etc): Solve[x^2 == 1, x, PositiveReals] {{x -> 1}}

One possibility is to mimic the behavior of symbolic constants like Pi, E, etc: N[x, _] ^= 4; NumericQ[x] ^= True; Then: Sqrt[x^2] x without even using Simplify.

The region of integration is: reg = ImplicitRegion[0 < y < a && 0 < x < z && 0 < z < y, {x, y, z}]; Using this region in Integrate: sol = Integrate[Exp[(a-x)^3], {x, y, z} ∈ reg, Assumptions -> a > 0] …

You could use SeriesCoefficient instead: Assuming[n ∈ Integers && n>1, Simplify @ SeriesCoefficient[ Exp[x] x^(-1-n), {x, 0, -1} ] ] 1/n!

Here's an unorthodox approach. Since $\int_{-\infty }^{\infty } f(x) \, dx$ is just FourierTransform[f[x], x, 0] (up to suitable FourierParameters), you could use UpValues to teach Mathematica about F …

Another method is to use PowerExpand: PowerExpand[Log[E^a], Assumptions -> True] a + 2 I π Floor[1/2 - Im[a]/(2 π)] Addendum As an aside, Log behave exactly like ArcSin here: ArcSin[Sin[x]] … Pi/2 < x < Pi/2] x For larger domains, Simplify doesn't work: Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi] ArcSin[Sin[x]] Again, using PowerExpand is useful: p = PowerExpand[ArcSin[Sin[x]], Assumptions …