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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


2

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]


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

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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