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8

df2 = D[df1, μ]; $Assumptions = Flatten[{Thread[{c1, c2, λ, μ} > 0], Element[{c1, c2, λ, μ}, Reals], μ > λ}]; FullSimplify@Positive[df2] (* True *) FullSimplify@Sign[df2] (* 1 *) Or, you can use your assumptions directly as the rhs of the Assumptions option, or as the first argument of Assuming, without setting the value of the global variable ...


4

You have a typo in your second equation. eqns = {2 + 2 a*d + 2 a*e == 0, 1 - 2 e + 2 d*b + 2 e*b == 0, 1 + 2 d*c + 2 e*c == 0, -2 + a^2 - 2 b + b^2 + c^2 == 0, -2 + a^2 + b^2 + c^2 == 0}; Solve[eqns, {a, b, c, d, e}] // Simplify


5

Oh, you just incorrectly type the equation in Mathematica, your second one should be: 1 - 2 d + 2 d b + 2 e b == 0, check the 2 d term. It's not a 2 b. Solve[{2 + 2 a d + 2 a e == 0, 1 - 2 d + 2 d b + 2 e b == 0, 1 + 2 d c + 2 e c == 0, a^2 + b^2 + c^2 - 2 == 2 b, a^2 + b^2 + c^2 - 2 == 0}, {a, b, c, d, e}] (* {{a -> -2 Sqrt[2/5], b -> 0, c ...


2

For me this whole thing remains rather mysterious: FullSimplify[-Sqrt[5 + 2*Sqrt[6]], ComplexityFunction -> LeafCount] gives the desired expansion despite the fact that SimplifyCount as per Chip Hurst's link SimplifyCount[-Sqrt[2] - Sqrt[3]] 17 shows a higher leaf-count than SimplifyCount[-Sqrt[5 + 2 Sqrt[6]]] 16 On the other hand ...


12

You need a custom ComplexityFunction. Essentially Simplify tries to minimize the SimplifyCount of the expression. This function is defined here. In your case the original expression is deemed simpler: SimplifyCount[-Sqrt[5 + 2*Sqrt[6]]] (* 16 *) SimplifyCount[-Sqrt[2] - Sqrt[3]] (* 17 *) Here's a custom ComplexityFunction: FullSimplify[-Sqrt[5 + ...


4

Here are three approaches to the function: f1[x_] := Piecewise[{{Sqrt[x^2], x != 0}}, 0] f2[x_] := Piecewise[{{Sqrt[x^2], x < 0}, {Sqrt[x^2], x > 0}}, 0] f3[x_] := Piecewise[{{-x, x < 0}, {x, x > 0}}, 0] Their second derivatives. Simplify@D[#[x], x, x] & /@ {f1, f2, f3} It seems that f2 or f3 might be used, but Simplify[f2[x]] ...


4

fun = Sqrt[x^2]; dd = D[fun, x, x]; With V10 we can explicitly define: {dd, FunctionDomain[dd, x]} Or {Simplify @ dd, FunctionDomain[dd, x]} {0, x < 0 || x > 0} Also with V10 you might consider Inactivate: di = D[Inactivate@Sqrt[x^2], x, x] // Together which prevents Together from evaluating to 0 di // Activate 0


6

Let's define f[x_] := Sqrt[x^2] FullSimplify[D[f[x],x], x ∈ Reals] (* ==> Sign[x] *) You can use f'[x] or D[f[x],x] interchangeably here. Then the second derivative is simply FullSimplify[D[Sign[x], x], x ∈ Reals] (* ==> Derivative[1][Sign][x] *) So doing the second derivative in two steps yields a mathematically correct answer. The ...


0

Here is a function to show the order of a list of expressions. By using Less as in the question, I'm assuming the expressions are unequal. ClearAll[showOrder]; SetAttributes[showOrder, HoldAll]; showOrder[{e__}] := Defer /@ Hold[e][[Ordering[{e}, All, Less]]] /. Hold -> Less OP's example: Block[{k = 1, m = 1}, showOrder[{Subscript[ω, 1], ...


6

I think this does what you want: Less @@ SortBy[Defer[Subscript[ω, #]] & /@ {1, 2, 11, 22}, N @@ # &]


1

You can use FunctionExpand for this. gammaFRatio[m_] := FunctionExpand[Gamma[n + m]/Gamma[n + 1]] Example: gammaFRatio[10] (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (8 + n) (9 + n)


7

x = E^-n (2 E^-1 + E^2) + E^n (2 E^-2 + E); x /. Times[a_, b_] :> Times[a, ExpToTrig@b] E^n (E + 2 Cosh[2] - 2 Sinh[2]) + E^-n (2 Cosh[1] + Cosh[2] - 2 Sinh[1] + Sinh[2])



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