# Tag Info

3

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_] = ...

3

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[#] & /@ ...

18

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 ...

0

Janis, I think the following approaches might be easier. I am going to name your expanded expression for convenience: expandedexpr = 1/2 Sqrt[3] Sqrt[(3 - mf) (5 + mf)] Ket[-(3/2), 3/2 + mf] - 1/4 Ket[-(1/2), 1/2 + mf] - 1/2 mf Ket[-(1/2), 1/2 + mf] + Sqrt[(4 - mf) (4 + mf)] Ket[1/2, -(1/2) + mf]; Easiest of all is to use Simplify on 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

0

This works nicely: PolynomialForm[(x + 1) (x + 2) (x + 3) // Expand, TraditionalOrder -> True] x^3 + 6 x^2 + 11 x + 6 The problem with PolynomialForm[] is that it is merely a wrapper, like MatrixForm[] for matrices. So, this makes things display pretty, but it should not be given as input for further calculations. You can copy the output, however.

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 + ...

4

As is well known, and has been discussed extensively in this forum, there may be problems in general with Integrate[] and the fundamental theorem of calculus, mostly due to discontinuities or other singularities in the antiderivative. But not in this case for version 8: \$Version (* Out[1]= "8.0 for Microsoft Windows (64-bit) (October 7, 2011)" *) The ...

4

I first suspected that the small imaginary component in the numeric approximation was the source of the problem but adding Re did not change the False result. Next I checked to see if the first fifty digits matched and they do: x1 = AppellF1[1/2, 1/2, 1/2, 3/2, 1/2 + I √3/2, 1/2 - I √3/2]; x2 = EllipticK[3/4]/2; Equal @@ RealDigits[{Re@x1, x2}, 10, 50] ...

6

This function lives in the system as Simplify`SimplifyCount.

3

Based on Vladimirs solution I wanted to post a faster alternative to SimplifyCount which produces the same results as SimplifyCount, but is a factor 3 faster. This can be very significant in case of complicated functions, it is however still significantly slower then Automatic. myNumberComplexity[x_Integer] := If[Positive[x], IntegerLength[x] - 1, ...

Top 50 recent answers are included