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1

My solution: expr = (a*x^2 + b*x*Sin[y] + c*Sin[y])^2 + (a*Sin[y]^2 + b*x)^3; (Cases[Collect[Expand@expr, x], x^2*__]/x^2) // First 2 a c Sin[y] + (1 + 3 a) b^2 Sin[y]^2

1

Mathematica has a built-in called Coefficient which solves the problem: Coefficient[(a*x^2 + b*x*Sin[y] + c*Sin[y])^2 + (a*Sin[y]^2 + b*x)^3, x, 2]

0

As stated in the comments, it is not possible to get the exact roots in the desired form. However, it is possible to get them in any arbitrary precision (100-digit precision in the following example): Rationalize[N[Solve[x^4 - 4 x^2 - 2 x + 1 == 0, x], 100], 0] {{x -> -1}, {x -> 148845339002531569051638627576397352071169969019092/ ...

0

From the documentation, $Assumptions is the default setting for the Assumptions option used in such functions as Simplify, Refine, and Integrate. It expects statements such as$Assumptions = a < 0 && b < 0}, using the wrong syntax $Assumptions = {n ∈ Integers, k ∈ Integers} actually breaks the use of Simplify. The correct approach should ... 0 Firstly, let's see this case: In[45]:= (FullSimplify[Sqrt[-a^2 - b^2], Assumptions -> #] &) /@ {{a > 0, b > 0, c > 0}, {a > 0, b < 0, c > 0}, {a < 0, b > 0, c > 0}, {a < 0, b < 0, c > 0}, {a ∈ Reals, b ∈ Reals, c > 0}} Out[45]= {I Sqrt[a^2 + b^2], I Sqrt[a^2 + b^2], I Sqrt[a^2 + b^2], I Sqrt[a^2 ... 2 In Mathematica, Sqrt[x]Sqrt[y]==Sqrt[x y] is not always true, there is a simple example gived by @Rahul Narain$\sqrt{-1}\times\sqrt{-1}=i\times i=-1\sqrt{(-1)\times(-1)}=\sqrt{1}=1$So FullSimplify[Sqrt[x] Sqrt[y]] don't give the result as Sqrt[x y] To get a better answer, to assume range of the variables is helpful, like this In[59]:= tps = (1/2) ... 2 Here is a way to do it, starting with the last expression in your question: Collect[ (4*(l^2 + m^2)* Pi/(1 + l^2 + m^2)^2 + ((Pi*(4*l* Derivative[0, 1][w][l, m] + (1 + l^2 + m^2)* Derivative[0, 1][w][l, m]^2 + Derivative[1, 0][w][l, m]*(-4*m + (1 + l^2 + m^2)* Derivative[1, 0][w][l, ... 2 Rather than a guess I could provide a reasonable explanation (when lacking it usually leads astray) thus we are prompting another way offering also understanding. Since the given equation is a functional one and Mathematica does not offer a direct functionality we have to deduce with the system an adequate scheme for solving such equations. Let's ... 5 As suggested by george2079 the equation can be solved by substition: Reverse@Simplify[f[x] == 2 x + 1/x - f[1/x]/2 /. f[1/x] -> 2/x + x - f[x]/2] f[x] == 2 x To find$f^{-1}(4)$you can use f = 2 # &; InverseFunction[f][4] 2 1 Although this is indeed a W|A question (take a look at the help center), here's several options for you. Inputting Simplify[-cos(n*pi)/(n*pi)] assuming n integer Gives this - link #1 -((-1)^n/(n Pi)) $$-\frac{(-1)^{n}}{n \pi}$$ As noted by @m_goldberg, inputting this also gives the same answer - link #2 simplify -cos (n*pi)/(n*pi) assuming n ... 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 + ...

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