Cubic equations and elliptic curves

At what ##c## does the curve change from one type to the other and what then is the leftmost point of the curve?

Answers$$c=\sqrt{\frac{32}{27}}\ \ ,\ \ x=-\sqrt{\frac{8}{3}}$$Here's why PF Challenge 2110#14.pdf (7 pages).

The Parallel Propagator and using it on the surface of a sphere

The three page Appendix I is on the parallel propagator which gives a general solution to the parallel transport equation of a vector. The appendix shows how to get this from the parallel transport equation which is, for a vector ##V^\nu## transported along a curve ##x^\mu\left(\lambda\right)## $$\frac{dx^\mu}{d\lambda}\partial_\mu V^\nu+\frac{dx^\mu}{d\lambda}\Gamma_{\mu\sigma}^\nu V^\sigma=0$$First we note that the transported vector can be calculated by$$V^\mu\left(\lambda\right)=P_{\ \ \rho}^\mu\left(\lambda,\lambda_0\right)V^\rho\left(\lambda_0\right)$$where ##V^\rho\left(\lambda_0\right)## is the vector at the start point ##x^\mu\left(\lambda_0\right)## and ##P_{\ \ \rho}^\mu\left(\lambda,\lambda_0\right)## is some matrix which is called the parallel propagator. Next we define another matrix $$A_{\ \ \rho}^\mu\left(\lambda\right)=-\Gamma_{\sigma\rho}^\mu\frac{dx^\sigma}{d\lambda}$$and then show that (dropping indices on the matrices)$$P\left(\lambda,\lambda_0\right)=I+\sum_{n=1}^{n=\infty}T_n$$where ##I## is the identity matrix and $$T_n=\int_{\lambda_0}^{\lambda}{\int_{\lambda_0}^{\eta_n}\int_{\lambda_0}^{\eta_{n-1}}{\ldots\int_{\lambda_0}^{\eta_2}A\left(\eta_n\right)A\left(\eta_{n-1}\right)\ldots A\left(\eta_1\right)}d^n\eta}$$$$=\frac{1}{n!}\int_{\lambda_0}^{\lambda}{\int_{\lambda_0}^{\lambda}\int_{\lambda_0}^{\lambda}{\ldots\int_{\lambda_0}^{\lambda}\mathcal{P}\left[A\left(\eta_n\right)A\left(\eta_{n-1}\right)\ldots A\left(\eta_1\right)\right]}d^n\eta}$$where ##\mathcal{P}## orders the matrices ##A\left(\eta_i\right)## so that ##\eta_n\geq\eta_{n-1}\geq\ldots\geq\eta_1##.

2-simplex

The first integral is over ##n##-dimensional equilateral right triangles, or ##n##-simplices and is quite hard to calculate but ##n!## ##n##-simplices make an ##n##-cube which makes the second integral which is much easier to calculate. I had a bit of trouble getting my head round all that and I tested it on a few examples including vectors transported along lines of constant latitude.  It all works!

Read about it here Commentary App I Parallel Propagator.pdf (13 pages). 

Vectors transported round latitudes calculated by parallel propagator

Note vector length never changes. ##T_n## was evaluated to 5 decimal places. 

Transport at 80°N. Vector barely
changes because it's nearly flat up there.
##T_n=0## at ##n=25##

Transport round equator. Vector remains
parallel because equator is geodesic.
##T_n=0## at ##n=1##

Transport at 15°N.
Vector rotates down by 87°.
##T_n=0## at ##n=11##

Transport at 15°S.
Vector rotates up by 87°.
##T_n=0## at ##n=11##