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3

Although it's not quite so obvious, Apart will always choose a variable when one isn't explicitly given. Apart[(a - b + c - d)/(a - b), c] $$\frac{c}{a-b}+\frac{a-b-d}{a-b}$$ vs Apart[(a - b + c - d)/(a - b)] $$\frac{a-b+c}{a-b}-\frac{d}{a-b}$$ and will choose in alphabetical inverse order which is the variable. (the last one in the alphabet. ) ...


0

Yet another possibility: ComplexExpand[Sqrt[2 + I], TargetFunctions -> {Re, Im}] // FunctionExpand // Simplify (5^(1/4) ((2 + I) + Sqrt[5]))/Sqrt[10 + 4 Sqrt[5]]


0

The identity works in a more broad case of $a\in\mathbb{C}$. Verification Simplify[a^(b c) - (a^b)^c, Assumptions -> b ∈ Integers && c ∈ Integers] (* 0 *)


4

For questions like these, FindInstance is your friend: FindInstance[a^(b c) - (a^b)^c != 0, {a, b, c}] (* {{a -> 99/5 + (12 I)/5, b -> -(28/5) + (79 I)/5, c -> 61/10 + (143 I)/10}} *) So at least for complex numbers, there are issues. We can try to ask for solutions with real values as well: FindInstance[{ a^(b c) - (a^b)^c != 0, (a | b | c) ∈...


0

You could introduce the perturbative part as δ[a] and define an extra rule how multiplication works with those (δ can be entered quickly via EscdeltaEsc or via \[Delta]): δ /: Times[___, _δ, _δ, ___] = 0; Now every time at least two δ[_] symbols are multiplied they will be simplified to zero automatically. For example in (* Input *) (a + δ[a]) (b + δ[b]) %...


3

I don't think that what you're seeing is a bug. The problem is that Sum implicitly assumes that the summation bounds are ordered and differ by an integer amount (mentioned here). The following example illustrates how this might cause issues: (* evaluate sum symbolically *) Sum[i, {i, a, b}] (* -(1/2) (-1 + a - b) (a + b) *) (* insert problematic values ...


2

I can't reproduce the first part you say. On V 12 on windows 10, I get 2.25 in both cases (it has to be, as simplifying 2.25 to 1.5 would be a really bad bug) But what I see, is that when changing the definition from delayed := to =, then now Mathematica gives 1.5 And this I think must be a bug. Code ClearAll[n, x, j, i, f, g]; f[n_, x_] := Sum[Sum[(-...


5

One way might be expr = Sqrt[-1]/a^(3/2); Assuming[a < 0, Simplify[ComplexExpand[expr]]] PowerExpand[%] Btw, you will not get the same exact expressions in both cases you showed. This is because $\sqrt{a^3} = a^{3/2}$ only for $a\geq 0$ and you said that $a<0$. ClearAll[a] Reduce[ Sqrt[a^3] == a^(3/2) && Element[a, Reals]]


8

Many times Mathematica gives enormous results to simple problems If Simplify still does not help reduce the antiderivative to what you like, you could always try Rubi << Rubi` integrand = 1/Sqrt[1 + Sin[x]]; sol = Int[integrand, x] D[sol, x] // Simplify There is a page here which compares different integrators with the size of antiderivatives ...


0

Setting the first statement equal to lhs and the second to rhs, we can ask for: FindInstance[lhs =!= rhs, {d, x, h}] {{d -> -(12/5) - I/2, x -> -(9/5) + (3 I)/5, h -> -(24/5) - (21 I)/5}} So they are not equal.


0

Rewriting problems can often be directly solved with replacements. Since there are a relatively small number of replacement rules involved, that would be my go to. I use ReplaceRepeated (//.) in dist below to implement these replacements, as there may be several layers of replacements to unravel as the distribution continues: dist[expr_] := expr //. {Dot[...


1

{1, I}.(Sqrt[2 + I] // ReIm // ComplexExpand // FunctionExpand // FullSimplify) (* I Sqrt[1/2 (-2 + Sqrt[5])] + Sqrt[1/2 (2 + Sqrt[5])] *) or % // Simplify (* (I Sqrt[-2 + Sqrt[5]] + Sqrt[2 + Sqrt[5]])/Sqrt[2] *) Verifying that these are equivalent to the original form Sqrt[2 + I] === (% // FullSimplify) === (%% // FullSimplify) (* True *)


0

Try Sqrt[2 + I] // ComplexExpand // FunctionExpand It may not create a result simplified in the exact form you want, but it will be closer.


1

One approach is to define your integral as a function, which you can then call with your desired x[s]: myHeavy[x_] := Integrate[HeavisideTheta[s - t] x, {s, 0, 1}, Assumptions -> {t > 0}]; Now, when you want to call this with a function x[s] = s^2, you evaluate myHeavy[s^2] To evaluate with Sin[s] myHeavy[Sin[s]] These seem to return plausible ...


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