Yeah, but the problem was something vague along the lines of: If you take circle tangential to a point p on the curve of smaller radius than the osculating radius then a small patch on the curve around p lies outside the circle
I didn't bother to make his problem precise. Here is my reformulation of his problem. Consider the normal map to the curve in R^2, given by f(t, s) = gamma(t) + s*N
It's in the spirit of (classical) algebraic geometry :P
You literally take the point and two other points on the curve; there's a unique circle through those. The limit of that circle as the two points collapse to the fixed point is the osculating circle. (Warm-up is tangent line as limit of secant lines.)
Similar to your exercise, @Balarka, is the observation that of all planes containing the tangent line to a space curve, only the osculating plane has the property that the curve lies (locally) on both sides of it.
That's kind of nonobvious, you'd think you can break that by taking something like a damped twisted cubic, so that one side lies above the plane and one side below
But no
I think the corresponding statement for the osculating circle is just false. Something like the graph of $x^2 - x^3$
Yeah, but you can't say anything about what happens to the curve at the osculating circle. I think the curve can lie on both sides of the osculating circle.
For radius less than $1/\kappa$ or greater than $1/\kappa$ you can
Locally, every curve looks like the twisted cubic (see Local Canonical Form). I edited my original claim. The curve is on both sides of the osculating plane; but it's the only plane with that property (containing tangent line).
How do you actually decide what reward value to give for each action in a given state for an environment?
Is this purely experimental and down to the programmer of the environment? So, is it a heuristic approach of simply trying different reward values and see how the learning process shapes up?
...
Power series are infinitely differentiable. Something without the first derivative is pretty hopeless.
If you do $|\cos x|$ it won't care, because near $0$ the cosine function is always positive. But you won't get any information outside of $[-\pi/2,\pi/2]$.
Speaking of osculating circles... I'm trying to understand the representation of curvature by the common convention $k = \frac{1}{R}$. What exactly does it mean to have $k$ curvature?
I was thinking, y'know, maybe there's a way to take a line segment that I can curve into an arc as part of a circle and in that way get arbitrary/linear rotation...
The minimal resolution for $V(I)$ at the origin can be described as follows; the exceptional divisors under iterated blowups are a bunch of $\Bbb{CP}^1$'s lying above $0 \in V(I)$. All of these have self-intersection number -2. The way these pairwise intersect are described by the Dynkin diagrams, according to the last column
Yeah, you can represent Poincare as surgery on something that looks like the Dynkin diagram. (Unknots for each vertex which are linked in order bases on the edges)
If you think of it instead as a handle diagram you get the E8 manifold w boundary P
