I understand that it can be shown that kinetic energy at speeds v<<c can be derived from the relativistic formula E=mc^2 to show the Newtonian KE=(1/2)mv^2. Hmmmm?
Well, yes, but you need another formula. The e=mc[sup]2[/sup] includes both kinetic energy and rest mass energy, because m is relativistic mass which gets bigger as the speed increases. The kinetic energy is ke=mc[sup]2[/sup] - m[sub]0[/sub]c[sup]2[/sup] where m[sub]0[/sub] is the rest mass of the particle. Relativistic mass is m[sub]0[/sub]/sqrt(1-v[sup]2[/sup]/c[sup]2[/sup]). Plug that into the ke formula and I think you can simplify it down to ke=1/2 m[sub]0[/sub] v[sup]2[/sup], if you take the limit as v/c << 1.
You need to use a Taylor’s expansion to reduce it further. Sqrt[1-x]~= 1-(x/2) for very small x. In this particular case, x = v^2/c^2. Substitute 1-(2v^2/c^2) for Sqrt[1-v^2/c^2] and you’ll reduce it down.