Comments by "EebstertheGreat" (@EebstertheGreat) on "The Goat Problem - Numberphile" video.

  1. FWIW, we had an exact answer from the start. It was the solution to π/2 = r² acos(r/2) + acos(1 - r²/2) - r/2 √(4 - r²). That solution can be found numerically, and it is "exact" in the sense that the solution to this equation also solves the original problem exactly; no approximations were made in the analysis. This is also exactly the kind of "easy answer" James is talking about with respect to polynomial equations. For instance, the equation x^5 - x - 1 = 0 has one real solution (and four distinct non-real solutions), but it cannot be represented by an elementary expression. The best way to write the answer is "the real solution to x^5 - x - 1 = 0." You can introduce new functions like the hypergeometric function to create an expression for this solution, but it's not a general method; if we move up to x^6 - x - 1 = 0, the solutions can no longer be represented with the hypergeometric function. There is no general way to solve these that is significantly better than just inventing a function that solves polynomial equations by definition. And this same problem arises in higher-dimensional versions of the goat problem. What we have now is a closed-form solution to the goat problem. It's no more exact than the original, nor is it any easier to compute. It still can only be found numerically and only to finite precision. So it's no more or less "exact," just a different way of writing it, but it's nice in that r can be represented with a single mathematical expression. For what it's worth, whether this actually counts as a closed form is also debatable, since expressions involving integrals are usually by definition not considered closed. Traditionally, a closed-form solution used only a finite number of operations. In fact, this is the first example I have found that describes an integral expression as a closed form. In any case, this is the first time anyone has successfully written any mathematical expression at all that exactly evaluates to the solution in question without making up new functions specifically for the problem at hand.
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  2.  @TheEternalVortex42  It's particularly disappointing when you realize that the process of evaluating these contour integrals amounts to solving the original equation but with extra steps.
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