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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 ...


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 ...


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


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


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 ...


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]


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 ...



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