ABSTRACT We consider a notion of excess curvature, which is simple in definition but quickly produces hard problems. Compare the arclength or curvature integral of a standard function plotted in a region. We can compare this to a flat line, and look at the difference across the entire integration domain. For probability distributions this produces a finite constant which is often unknown. INTRODUCTION If the arclength of a curve f(x) in a region x ∈ [a, b] is given by $$ _a^b[f] = \int_a^b {\partial x}\right)^2} \; dx $$ and for the curve f(x)=c, for some constant, we have 𝒜ab[c]=b − a we can consider the _excess_ curvature as the difference 𝒜ab[f]−𝒜ab[c] which is curvature beyond the domain range. As an integral this is simply $$ _a^b[f] = \int_a^b {\partial x}\right)^2} -1 \; dx $$ the reason for doing this is to extend to an infinite (a → −∞,b → ∞) or semi-infinite domain, where the individual 𝒜 terms diverge, but the difference is convergent. We can then remove the limits a and b, as long as we are clear on the support being integrated over. EXAMPLES Consider the unit Gaussian $$ f(x) = {} e^{-x^2/2} $$ we then have ℬ[f]=0.069796988349688... which does not immediately have a clear closed form. If we shift the mean of this distribution it has no effect on the final result, as the hump is the same distortion to the line as before. We can get closed forms for a few very simple distributions, for example the unit triangle t(x) and it’s normalised powers which give $$ [t(x)] = 2 - 2 \\ \left[{2}t^2(x)\right] = +(3) - 2 \\ \left[2t^3(x)\right] = }{3} -2 - {3}F(i((1+i)),-1) $$ where F is the elliptic F function, showing how complicated these closed forms get. Because the support of these is always [ − 1, 1], there is always a term of −2 which is the base curvature of that support. For infinite support it is less clear how this factor will look. For a function which is not a probability distribution, such as f(x)=ex, the excess curvature in the infinite limit is unbounded. But we can find closed forms for ^y_{-\infty}[e^x] = -(y+1) + } + \left(}}{3-5 }}\right) it is apparent that the term −(y + 1) looks like the support term, and is separable even in this semi infinite case. On the range [0, ∞), we have \left[}{2}\right] = }{2} -1 - (2) + \log(4) on the symmetric full support for which this is a normalised distribution we can double this as \left[}{2}\right] = -2 - 2(2) + 4\log(2) In general it seems on the range [0, ∞), we have \left[}{2}\right] = -{a} + }{2a} - {a}\left({a^2}\right) + {a} - {a} COMBINATIONS For combinations of unit triangles t(x) we have non-additive effects. $$ t(x) \to (2-2) \\ {2}(t(x-\delta)+t(x+\delta)) \to (2 - 4), \delta \ge 1 $$ in between this, there is some kind of overlap function, where the curvature of the combination is different to the pair or the individual. $$ 2 \left(-2\right) & d>1\lor d\leq -1 \\ -2 \left(-2\right) d & -1<d\leq -{2} \\ 2 \left(-2\right) d & {2}\leq d\leq 1 \\ 2 \left(-1\right) (2 d+1) & d=0 \\ -2 \left(- d+2 d-+1\right) & 0<d<{2} \\ 2 \left(- d+2 d+-1\right) & $$ for three triangles fully separated we have $2 - 6$, for four we have $2 -8$. In general it seems $2-2n$ for n non-overlapping triangles scaled by 1/n. If the triangles are not scaled, then we have additive curvature as expected to give $2n(-1)$. We could potentially see this as a form of convolution, excess curvature convolution, for a pulse f(x) we measure $$ [f * f](t) = ^\infty -1 \; dx $$ or with averaging $$ [f * f](t) = ^\infty {4}\left(f'(x)+f'(x-t)\right)^2}-1 \; dx $$ WHY DO THIS? Is there an ’energy’ associated with curvature? Consider a probability density of a particle or wavepacket, or a deformation in a surface such as a wave. What about constructive and destructive interference? Consider calculus of variations trying to minimise a quantity. If we take an ensemble of particles each with Gaussian density, what is the minimum curvature configuration? For the triangular function above, the minimum comes with (t(x)+t(x − 1))/2. This brings the particles together but does not overlap them and does not push them too far apart. What is the forcefield associated with such dynamics? Can we define a potential energy function V(x) or pairwise interaction which has the same minimum and gradients? Upgrading to a Gaussian means the tails will interact over distance. For a unit Gaussian, the equilibrium point lies somewhere between a shift of 2.447 and 2.451, another minimum is full separation of the two bodies.
