I said
b) although the equation of motion of an elliptical orbit requires higher transcendental functions,[...] (The hard part is theta(t), which apparently even transcends Lambert_W.)
Solving Kepler's equation, E - epsilon sin(E) = M, for E also solves the old math teacher embarrasser of finding the thickness x of the circular segment of given area a, e.g., 2 acos(1 - x) - sin(2 acos(1 - x)) = 2 a. The analytic solution is inf ==== sin(2 a n) Bessel_J (n) \ n x = 1 - cos( > ----------------------- + a). / n ==== n = 1 E.g., to slice a unit disk into four equal areas using parallel cuts requires "heel" thickness inf n ==== (- 1) Bessel_J (2 n - 1) %pi \ 2 n - 1 x = 1 - cos(--- - > -------------------------------) ~ 0.5960 . 4 / 2 n - 1 ==== n = 1 Convergence is lousy and the big Bessels have bad cancellation--you're way better off with Newton's method. Alternatively, revert the a(x) series (c192) sphere_segment_volume(n,1,x) n - 1 n - 1 n + 1 ----- ----- ----- 2 2 n + 1 1 - n n + 3 x 2 2 %pi hyper_2f1(-----, -----, -----, -) x 2 2 2 2 (d192) -------------------------------------------------------- n + 1 (-----)! 2 with (dimension) n = 2, yielding 2 3 4 5 6 b 11 b 823 b 150653 b 3362377 b (d294)/T/ x = b + -- + ----- + ------ + --------- + ---------- + . . . 10 350 63000 24255000 1051050000 2 9 a 1/3 with b = (----) . 32 (x ~ area^(2/3) ?!) Or you could even use the "continued sin" iteration 2 a + sin(2 a + sin(2 a + sin(2 a + . . .))) x = 1 - cos --------------------------------------------. 2 For the thickness of a spherical segment, you need only solve the cubic x 2 v = pi (1 - -) x . 3 E.g., a depth of pi x = 1 - 2 sin -- ~ 0.6527 18 quarter-fills a spherical tank of unit radius. Apropos the cyclotomic discussion, I assume everybody knows about the 2s in factor(x^105-1). What is the lowest degree polynomial with coeffs in {-1,0,1} with a coeff of 2 in its irreducible factorization? --rwg PS (reminder): a really obscure typo on p 573 of Watson's Bessel treatise: oo ==== \ 2 1 1 (2) > Bessel_J (n z) = -------------- - - / n 2 2 ==== 2 sqrt(1 - z ) n = 1 is OK, (although the summand looks more like Bessel_J (n z) 2 n ) but the rhs of oo ==== 2 2 \ 2 2 z (z + 4) (3) > n Bessel_J (n z) = -------------- X X / n 2 1/2 <<< X ==== 16 (1 - z ) X X n = 1 should have been the much less plausible 2 2 z (z + 4) --------------. 2 7/2 16 (1 - z ) Luckily TeX, though written at an AI lab, contains no AI. (Actually, really good AI would flag the inconsistent sqrt notations in the denominators, or even test the formulas!)