I'll fiddle with that MathJax bookmark and then get back to it.

@mathFromtheGroundUp yeah, that was me

put the mathjax link in your boomkarks, make sure to have your bookmarks toolbar showing so you can actually click the bookmark

then be in this tab in your browser and click it

it should turn the following into math: $\Bbb C[G]\cong\bigoplus V\otimes V^*$.

Okay, got it.

Let me parse #3...

"At every point on a curve there is a unique tangent line." At that point t, there may be other points on the curve with that magnitude of tangent.

"This line is the best local approximation of the curve to the first order." First derivative, in other words.

"But a line can be thought of as a generalized circle, one of infinite radius and whose center exists "at infinity," so perhaps consider all circles that are tangent to the curve at a point - which is the best fit?"

So you can "zoom in" on a piece of a circle to make it look like a line and conversely "zoom out" on some part of a curve to see the part of the curve that could be approximated by a circle

So from the examples in UC Davis the "best fit" circle at a point looks to be a circle which "fits" on the concave side of the part of the curve you're interested in finding the curvature for.

So a circle that is just large enough to fit inside the part of the curve you're interested in?

[ossculating circle][1] [1]: mathwiki.ucdavis.edu/@api/deki/files/702/…

hmm image posting isn't quite right in here.

Well that does link correctly at least. =/

"Which approximates it best to the second order?" So the second derivative.

Do you mean the rate of change of the radii of the circles as we traverse along the curve? Or the rate of change of direction of the tangents we talked about?

Maybe you mean fitting a circle to the graph of the first derivative?

@mathFromtheGroundUp if a curve and its tangent line are both understood as graphs of functions, then they share the same first-order taylor polynomial approximation at the relevant point. if we understand the osculating circle as the graph of a function too, then that function shares the same second-order taylor polynomial

@mathFromtheGroundUp the osculating circle can intersect the curve at other points, but locally that sounds like it might be right