# Tag Info

21

I've decided to expand on my comment. Before I delve into the solution, let's all pause for a moment and marvel at the stereographic parametrization of a unit circle: $$\begin{pmatrix}\frac{1-t^2}{1+t^2}\\\frac{2t}{1+t^2}\end{pmatrix}$$ Sometimes also referred to as the Weierstrass substitution, it has often been used as a tool in the solution of algebraic ...

9

Is this what you want? Clear[Derivative, h]; h[0] = 1; (* to avoid division by zero with OP's example R *) Derivative[1][h][s_] := Block[{Derivative}, h'[\[FormalS]] /. First@Solve[ D[R[a + h[\[FormalS]]] == - \[FormalS], \[FormalS]], h'[\[FormalS]]] /. \[FormalS] -> s ]; Derivative[n_][h][s_] := D[Derivative[n - ...

6

Update Based on Guesswhoitis. answer I have improved my ugly code and use his approach. Otherwise the format is as outlined in original answer. Manipulate[p = {-a, 0}; q = {0, b}; r = {c, 0}; s = mp[a, b, c]; nfb = RegionNearest[Circle[{0, 0}, b]]; nfc = RegionNearest[Circle[{0, 0}, c]]; res = VectorAngle @@@ Partition[Join[{{-a, 0}}, sc[#] & /@ ...

6

MyLine is embedded in 2D space, thus for $x\in\text{MyLine}$, x is a 2D point and Sin[x] just makes no sense. You probably meant the interval Interval[{-Pi,Pi}] instead, which I think should work. But it doesn't. I don't know why. Maybe a bug? In[27]:= MaxValue[Sin[x], x \[Element] Interval[{-Pi, Pi}]] During evaluation of In[27]:= MaxValue::objfs: The ...

6

(-1)^(1/3) (-Log[x])^(2/3) + Log[x]^(2/3) // FullSimplify[#, x > 1] & 0 Alternatively, using the real-valued cube root of x CubeRoot[-1] CubeRoot[(-Log[x])^2] + CubeRoot[Log[x]^2] 0 CubeRoot[-1] CubeRoot[-Log[x]]^2 + CubeRoot[Log[x]]^2 0

5

Since "Visualizing the resulting triangle is left as an exercise for the interested reader" and I am interested here is a visualization of J.M.'s numeric solution. DynamicModule[{corners, perimeter, sol, u, v, pts}, Manipulate[ corners = {{-c, 0}, b {(1 - u^2)/(1 + u^2), 2 u/(1 + u^2)}, a {(1 - v^2)/(1 + v^2), 2 v/(1 + v^2)}}; perimeter[u_, v_] = ...

2

Line defines a 2D region, therefore MyLine = Line[{{-Pi, 0}, {Pi, 0}}]; MaxValue[First@Sin[{x1, x2}], {x1, x2} ∈ MyLine] or MaxValue[Sin[First@{x1, x2}], {x1, x2} ∈ MyLine] or MaxValue[Sin[Indexed[x, 1]], x ∈ MyLine] would be the correct syntax. But it's much simpler to use MaxValue[{Sin[x], -Pi < x < Pi}, x]

2

Since I complained that the result returned by Mathematica is not as simple as I would like, I might as well post the closed form that I have. I will not write the derivation here, but the procedure is similar to what I did in this math.SE answer: N[InverseJacobiCN[-1/3, 3/4] - EllipticK[3/4], 20] 0.64617199330515618196 NIntegrate[1/Sqrt[x (1 - x + ...

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