A nice proof of the fact that the Quadratic Term of the Taylor polynomial is Curvature For plane curves, twice the quadratic term of the Taylor expansion at a point on that curve is precisely the curvature. I know of one proof as featured in Hubbard and Hubbard, but I was wondering if there was another more elegant solution.
Thanks!
 A: I'm unfamiliar with your reference, so please forgive me if this might be reminiscent of its solution.  I'll use the Serret-Frenet formulae.  For a twice differentiable curve $\gamma$ parameterized by arclength, one of them asserts that the normal component of the derivative of the (unit) tangent vector is the curvature $\kappa$.  (This is often taken as the definition of curvature.)  To simplify things, but wlg, let's use coordinates in which the point in question is the origin, so that the Taylor expansion is
$\gamma(t) = \left(a_1 t + \frac{a_2}{2}t^2 + \cdots, b_1 t + \frac{b_2}{2}t^2 + \cdots \right)$.
Whence the unit tangent vector at the origin is $T = (a_1, b_1)$ and its derivative is $(a_2, b_2)$.  To obtain the component normal to $T$, it suffices to compute the inner product of this derivative with the rotated tangent vector $(-b_1, a_1)$ (because the tangent vector has unit length), yielding
$\kappa = -a_2 b_1 + b_2 a_1$.
This general result, specialized to the case where $\gamma$ is of the form $(t, f(t))$ for a function $f(t) = 0 + b_1 t + \frac{b_2}{2}t^2 + \cdots$, reduces (because $a_1 = 1$ and $a_2 = 0$) to $b_2$, "twice the quadratic term in the Taylor expansion" at this point.
A: Here's a completely different answer inspired by J.M.'s description of the text.  It might give a bit more geometric insight.  It essentially shows that (a) twice the second coefficient in the Taylor series describes a parabola closely approximating the curve at the origin, (b) the curvature is determined by a circle closely approximating the curve at the origin, and (c) the curvature of the Taylor series parabola is twice that coefficient.  In other words, when we're talking about curvature we're really talking about the second-order behavior of the curve and that's exactly what the quadratic term in the Taylor series describes.
The relevant definition of the curvature here is the reciprocal of the radius of the osculating circle (taken to be negative if the circle lies below the curve, with 1/infinity = 0).  An osculating ("kissing") circle is found by taking three points close to the point under consideration (which again might as well be the origin), fitting a circle to them, letting the points independently converge to the origin, and seeing what the circle settles down to.  Assuming the circle exists, we can find it by considering some special configurations of points.  Choose the three consisting of the origin itself, with coordinates (0,0), and the points at nonzero distance $x$ to both sides of the origin.  Their coordinates are given by the Taylor series.  One is is at $(x, \frac{b}{2}x^2 + O(x^3))$ and the second is at $(-x, \frac{b}{2}(-x)^2 + O((-x)^3))$.  The center of the circle is equidistant from all three.  If we choose $x$ small enough (witness some geometric hand-waving here that can be made rigorous at the expense of making the following algebra a little harder), the situation is essentially symmetric around the origin and we expect the center to be located directly above (or below) the origin at a point $(0,a)$.  Consider the distances from $(0,a)$ to the origin and to the point corresponding to $x$: their squares must be equal.  Whence
$a^2 = (x-0)^2 + (\frac{b}{2}x^2 - a)^2 + O(x^3)$,
at least to an excellent approximation when $x$ is small.  Some terms cancel and an easy algebraic simplification gives (remember, we're solving for $a$ in terms of $x$ and $b$)
$0 = 1 - a b + \frac{b^2}{4}x^4 + O(x)$
implying $1/a = b$ in the limit as $x \to 0$.  Since $1/a$ is the curvature, our task is done.
Yes, as I have warned, this lacks rigor, but the rigor can be supplied (I recall Michael Spivak does this early in Volume II of his Differential Geometry series) and what it lacks in sophistication may be compensated by the elementary geometric intuition it affords of both curvature and the Taylor series.
