# Tag Info

9

The identity does not hold for x < -1: Plot[ProductLog[x*Exp[x]], {x, -5, 5}] FullSimplify[ProductLog[x*Exp[x]], x >= -1] x (* result in 10.1.0 under Windows *)

8

Just avoid the branch cut: Refine[Sqrt[Exp[I p]], π > p > 0] (* ==> E^((I p)/2) *) The branch cut of the square-root function doesn't allow Mathematica to simplify the expression without the restriction on p from above. See also the related question and answer here: Using Solve to solve the equation x1/3=−1

6

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[...

4

The equation x Exp[x] == y has multiple solutions for x. For example, evaluating tab = Table[{x -> ProductLog[i, 1]}, {i, 0, 5}] Exp[x] x /. tab N[tab] gives {1, 1, 1, 1, 1, 1} and {{x -> 0.567143}, {x -> -1.53391 + 4.37519 I}, {x -> -2.40159 + 10.7763 I}, {x -> -2.85358 + 17.1135 I}, {x -> -3.16295 + 23.4277 I}, {x -> -3....

4

PowerExpand[ProductLog[x Exp[x]]] x This assumes $x\ge0$

3

Using an undocumented function: Cosh[Sqrt[-a^2 - b^2]] /. Sqrt[expr_] /; Internal`SyntacticNegativeQ[expr] :> I Sqrt[-expr] Cos[Sqrt[a^2 + b^2]]

3

In these cases, you need to specify the variables are $>0$. Of course it also works for $<0$, but it just works this way in the software. FullSimplify[Cosh[Sqrt[-a^2 - b^2]], {a > 0, b > 0}] Cos[Sqrt[a^2 + b^2]] FullSimplify[Sqrt[-a^2 - b^2], {a > 0, b > 0}] I Sqrt[a^2 + b^2]

2

As usual, GroebnerBasis[] is very useful here: det[aa_] := Det[Table[x[li1, li2], {li1, aa /@ {1, 2, 3}}, {li2, aa /@ {4, 5, 6}}]]; origExpr = det[a] det[b]/det[c] + det[d] det[e]/det[f] - det[g] det[h]/det[i]; oeet = origExpr // Expand // Together; varsx = Cases[oeet, x[a_[i_], _], ∞]; vars = Union[Cases[oeet, x[a_[i_], _] :> a, ∞]]; rep = Table[d[ti] ...

2

Here's a possible approach: Clear[a, b, c, s, int] a = 4 Cos[af/2]^2; b = 1 - 5 Cos[af/2]^2; c = Cos[af/2]^2; s = Sqrt[a*x^4 + b*x^2 + c]; int[par_?NumericQ] := With[{integrand = s /. af -> par}, NIntegrate[integrand, {x, 0, 1}]] Plot[int[af], {af, -2 Pi, 2 Pi}] Finding the maxima and minima using the int expression is also possible. Let's first ...

2

Just give (Full)Simplify its assumptions in the second argument, for example: Simplify[u[x,y,z,t], {x > 0, y > 0}] (* -2 x + ((-E^x^2 + E^y^2)*y)/(E^x^2*Sqrt[x^2 + y^2]) *)

2

Try this: expr1 = (-2 Sqrt[x^2 + y^2] - ((1 - E^(-x^2 + y^2)) y)/x)/ Sqrt[1 + y^2/x^2]; then expr2 = MapAt[HoldForm, expr1, {2, 1, 2, 2}] // Simplify[#, {x > 0, y > 0}] & // ReleaseHold (* (-y + E^(-x^2 + y^2) y - 2 x Sqrt[x^2 + y^2])/Sqrt[x^2 + y^2] *) Take care that the assumptions x>0 and x<0 give different results. Have fun!

2

This is not an answer, but here's some analysis of the issue which didn't fit in a comment. First, let us rewrite your assumptions in a slightly different form: asmp = And @@ {0 < Ijm1 < Ij < Ijp1 < 1, bpjm1 < 0, bpj < 0, bpjp1 < 0, Ijm1 ∈ Reals, Ij ∈ Reals, Ijp1 ∈ Reals, bpjm1 ∈ Reals, bjp ∈ Reals, bpjp1 ∈ Reals}; Now ...

1

I figured out how to update this function to correctly work on multi-linear expressions. All that's required is one additional call of PolynomialReduce in the third nested If statement. This mirrors the way the second If statement handles the power case. It might be that there should be a similar treatment for the division case (the first If statement), but ...

1

The key idea of this solution is to partly simplify the expression by direct substitution of some known terms rather than using functions for simplification like Simplify. First, take out the imaginary part of the expression and observe: impart = E^(I ω t) Γ[I ω] + E^(-I ω t) Γ[-I ω] // Im // ComplexExpand Hmm… seems that there're many Arg terms in ...

1

I. A summary for the failing trial Residue can't calculate the residue of u directly. (It's hard to tell whether it's a bug or not, Residue never promises he will calculate every known residue, anyway.) SeriesCoefficient won't give desired answer in the following case: SeriesCoefficient[u, {s, 0, -1}] If it gave the desired answer, we would be able ...

1

Just trying to evaluate the function fun a few times on some different y1 and x1 values like fun/. y1 -> 3 /. x1 -> 2 // Simplify ((-4 + x2) (1 + 2 u + x2))/((-3 + x2) (2 u + x2)) One can infer that the whole hypergeometric function actually in general reduces to just ((-1 + x2 - y1) (-2 + 2 u + x2 + y1))/((-1 + x2 - x1) (-2 + 2 u + x1 + x2)) ...

1

I think it will work much faster if you define a domain for a,b,c. Otherwise you will end up with a lot of Conditionals. for example con1 = (2*a/x >= a/x + c + (1/(x))*Sqrt[a^2*(u^2 + 1) - 2*a*(b*u + c)*x + (b^2 + c^2)*x^2]) con2 = (2*(u*b + c)/(u^2 + 1) >= a/x + c + (1/(x))* Sqrt[a^2*(u^2 + 1) - 2*a*(b*u + c)*x + (b^2 + c^2)*x^2])...

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