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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 + ...


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 ...


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])


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 ...


6

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


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 ...


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


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


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 ...


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)


1

Obviously what a person thinks is simplified and what a given CAS thinks is simplified will not always be the same. This probably happens more often with functions like the trigonometric functions that admit so many identities. Often it does not matter that much, but when it does, a certain amount of personal intervention may be required. One hump to get ...



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