(sorry accidental misfire, my dominant wrist is injured) Continuing... If you take the integrals: Integrate[Hypergeometric2F1[1/2, 1/2, 1, x], {x, 0, 1}] == 4/Pi Integrate[Hypergeometric2F1[1/2, 1/2, 1, 1 - x], {x, 0, 1}] == 4/Pi And compare with Harold Edwards' normal form for elliptic curves, then it is easy to interpret the integrals as the area 4 of a 2x2 square, divided by Pi. Transformation of the elliptic curve explains why the quartic lemniscate along one complex dimension would again have a simple area integral. You could then take the geometry as a diversion to physics or even these days to Elliptic Curve Cryptography. Instead, how about a few more maths curiosities: Integrate[Hypergeometric2F1[1/2, 1/2, 1, x^2], {x, 0, 1}] ==4*Catalan/Pi Integrate[Hypergeometric2F1[1/2, 1/2, 1, 1 - x^2], {x, 0, 1}] == Pi/2 There is no digital pattern, nor any pattern in continued fraction digits, but both of these numbers have relatively simple Dirichlet series. Compare with: Integrate[Hypergeometric2F1[1/6, 5/6, 1, x^2], {x, 0, 1}] ==3*Sqrt[3]*ArcCsch[Sqrt[2]]/Pi Integrate[Hypergeometric2F1[1/6, 5/6, 1, 1 - x^2], {x, 0, 1}] ==3*Sqrt[3] /4 So, hmm... the z=1-x^2 integrals differ--only one algebraic--while the x^2 integrals are similar with a factor 1/Pi. If we follow Bruce Berndt and define signature s, then rewrite parameters a=1/s, b=(s-1)/s, c=1, then we have Elliptic K for s=2, and Ramanujan K for s=3,4, and 6. Elliptic curves are well known for z=x, z'=1-x. When we go to quadratic argument: z=x^2, z'=1-x^2, then the aforementioned integrals determine areas interior to the following few lemniscate curves: Show[ ContourPlot[ p^2 - q^2 + (4/27) (q^6) == 0, {q, -2, 2}, {p, -2, 2}, ContourStyle -> Red], ContourPlot[ -p^2 + q^2 - (4/27) (-3 q p^2 - q^3)^2 == 0, {q, -2, 2}, {p, -2, 2}, ContourStyle -> Blue ] /. {x_, y_} :> {x, y + 4}, ContourPlot[ p^2 - q^2 + (1/4) (q^4 + 6 q^2 p^2 + p^4) == 0, {q, -2, 2}, {p, -2, 2}, ContourStyle -> Green ] /. {x_, y_} :> {x, y + 2}, PlotRange -> {{-2.5, 2.5}, {-1.5, 5}}] https://0x0.st/zcZ0.png At this point we can start to distinguish between Elliptic K and Ramanujan K. Why is it so easy to find an algebraic model for the Ramanujan integrals with quadratic argument? Does an algebraic model exist for Elliptic K with quadratic argument? This last question seems huge, because it would give another (possibly new) definition of Catalan's constant. Or focusing on R's three chosen parameter sets, why should the periods on higher-genus curves work out to satisfy such a simple D.E.? The situation is similar to what we have seen with the Klein Quartic, just one curve. But here we have energy modulus z and three entire families!
From reading the archives, I see that N O A M E L K I E S sometimes writes the list. His article "Klein Quartic in Number Theory" covers similar ground, and seems to show a sense of fun for symmetry analysis and period integrals.
So why shouldn't Noam Elkies respond with a perspective on the decomposition symmetries of Riemannian surfaces? Or is this request a geopolitical no-go? Or a class issue? Because I'm new? Or what? According to the "Man who Knew Infinity", near where Ramanujan meditated for three days to obtain adesh: "the air was full of smoke, and the stone walls black with incense". What a nice detail giving a sense of time, where three days does not seem like too long to wait! The history around these integrals is also quite fun and enjoyable to learn! Cheers, Brad On Mon, Apr 29, 2019 at 1:33 PM Brad Klee <bradklee@gmail.com> wrote:
You know, the Fibonacci numbers are really fun to study, especially in relation to the golden mean, Penrose tiling, golden spiral, etc. Analysis via Benford's law, though somewhat interesting, unfortunately makes the Fibonacci idea look terribly one sided.
Now if you take the integrals
On Sat, Apr 27, 2019 at 6:37 PM Victor Miller <victorsmiller@gmail.com> wrote:
Look at this: https://mathoverflow.net/questions/87692/leading-digits-of-fibonacci-sequenc...