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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!

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I got confused since what you said was a bit different from how I knew curvature, but if you look at the book, the first treatment was to rotate axes such that the tangent was horizontal and the normal is vertical. Only then does "twice the quadratic term of the Taylor expansion at a point on that curve is the curvature" apply. Of course, under arclength parametrization, things become even simpler. –  J. M. Sep 16 '10 at 4:03
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For the curious: the formula for the tangential angle $\varphi$ (the angle that the tangent to a curve $\mathbf{r}(t)=(f(t)\;g(t))$ at $t=t_0$ makes with the horizontal axis) is $\varphi=\arctan\left(\frac{g^{\prime}(t_0)}{f^{\prime}(t_0)}\right)$ ; the rotation matrix needed to rotate the tangent to the horizontal position is then easily derived, since $\cos\;\varphi=\frac{f^{\prime}}{\sqrt{f^{\prime 2}+g^{\prime 2}}}$ and similarly for $\sin\;\varphi$. Only then do you Taylor-expand the rotated curve. –  J. M. Sep 16 '10 at 4:23
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On the other hand, the approach I was first introduced to was to consider the Argand form of the plane curve $z(t)=f(t)+i\;g(t)$ ; the required rotation is a multiplication of $z(t)$ with the factor $\exp(i\varphi)=\sqrt{\frac{z^{\prime}}{\overline{z^{\prime}}}}$ –  J. M. Sep 16 '10 at 4:27
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2 Answers

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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.

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This is fine, thanks. Note J.M.'s comment above, it is relevant here. –  BBischof Sep 16 '10 at 4:12
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For BBischof: later in the book, in the section on space curves, the equations for the Frenet frame (a.k.a. the Frenet-Serret equations) are displayed; this can be specialized to the plane curve case by setting the torsion $\tau=0$. –  J. M. Sep 16 '10 at 4:14
    
To further add: the nice thing about Frenet-Serret is that due to this, the Cesàro equation of a (plane) curve (an equation that relates the curvature and arclength) completely determines the curve except for position/orientation. (Positioning the initial Frenet frame appropriately will then determine position/orientation.) –  J. M. Sep 16 '10 at 5:25
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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.

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It may not be rigorous, but I sure like this way of looking at it. +1. –  J. M. Sep 16 '10 at 5:17
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