Definition: Let $M$ be a smooth $d$-manifold embedded in $\R^n$ and $N^\epsilon_pM = N_pM \cap B(p,\epsilon)$ the natural embedding of the $\epsilon$-normal plane at $p\in M$. The pairwise conditioning number of $p$ and $q$ is
\[
\tau_{p,q} = \sup\{\epsilon \ :\ N^\epsilon_pM\cup N^\epsilon_qM\text{ embeds in }\R^n\}.
\]
The condition on $\epsilon$ is the same as saying $i(N^\epsilon_pM)\cap i(N^\epsilon_qM) = \emptyset$, where $i$ is induced by the embedding of $M$. It is immediate that $\tau = \inf_{p,q}\{\tau_{p,q}\}$, so we will try to find $\tau_{p,q}$ first. Recall that a helix of radius $r$ and vertical period $2\pi c$ is a 1-dimensional manifold
embedded in $\R^3$ as the zero locus of
\[
f(x,y,z) = x-r\cos(z/c),
\hspace{2cm}
g(x,y,z) = y-r\sin(z/c).
\]
We first find the normal plane at two arbitrary points $p_1,p_2$ on the helix, then their intersection (which is a line), and then the distance from $p_1$ and $p_2$ to that line. The smallest of these two distances bounds $\tau_{p_1,p_2}$ from below (and the bound is achieved on pairs of points defining the medial axis). Then take the infimum of this value over all points on the helix. However, this excludes the case when the normal planes are parallel (for instance when the two points have the same $x$- and $y$-values).
Moreover, even just calculating the infimum for points whose normal planes are not parallel yields a result of zero. We describe the process nonetheless. For the first step, we need the equations of the normal planes. Let
\[
D^f = \begin{bmatrix} 1 & 0 & r\sin(z/c)/c \end{bmatrix},
\hspace{2cm}
D^g = \begin{bmatrix} 0 & 1 & -r\cos(z/c)/c \end{bmatrix}.
\]
be the Jacobians of $f$ and $g$. The points $p_1$, $p_2$ are completely described by the $z$-coordinate, so we have two values $z_1$, $z_2$ for $p_1$, $p_2$, respectively. The normal plane at $p_i$ is the zero locus of
\[
\det\begin{bmatrix}
x-r\cos(z_i/c) & y-r\sin(z_i/c) & z-z_i \\ 1 & 0 & r\sin(z_i/c)/c \\ 0 & 1 & -r\cos(z_i/c)/c
\end{bmatrix} = z - z_i - \frac{xr}c \sin(z_i/c) + \frac{yr}c\cos(z_i/c).
\]
We have two equations and three unknowns, so one independent variable. Solving for $x$ and $y$ gives us
\[
x = \frac{(z-z_1)\cos(z_2/c) - (z-z_2)\cos(z_1/c)}{r\sin(\frac{z_1-z_2}c)/c},
\hspace{1cm}
y = \frac{(z-z_1)\sin(z_2/c) - (z-z_2)\sin(z_1/c)}{r\sin(\frac{z_1-z_2}c)/c}.
\]
These are functions of $z$, giving us two new functions
\[
h_i(z) = (x(z)-r\cos(z_i/c))^2+(y(z)-r\sin(z_i/c))^2 + (z-z_i)^2,
\]
for $i=1,2$, which, when minimized, give a lower bound for the pairwise conditioning number of $p_1$ and $p_2$. Indeed, by slowly increasing the $\epsilon$ until the $\epsilon$-normal planes at $p_1$ and $p_2$ intersect, the first point of intersection will happen on the intersection $N_{p_1}M\cap N_{p_2}M$. Hence finding the shortest distance from $p_1$ and $p_2$ to this line gives a definite lower bound. The functions $h_i$ are quadratic in $z$, and we know the function $az^2+bz+c$, for $a>0$, has minimum at $-b/2a$. The values of $h_1$ and $h_2$ at their minima are the same and equal to
\[
h_m := h_i\left(\frac{-b}{2a}\right) = \frac{2(c^2+r^2)\cos^2\left(\frac{z1-z2}{2c}\right)\left(r^2+c(z_1-z_2)\csc\left(\frac{z_1-z_2}c\right)\right)^2}{2c^2r^2+r^4+r^4\cos\left(\frac{z_1-z_2}c\right)}.
\]
A natural limit of $h_m$ to consider is $z_2\to z_1$. If any of the factors in the numerator are zero, we also get a minimum, so another limit to look for is $z_2\to cz_1$, which makes the cosine factor zero. These are
\[
\lim_{z_2\to z_1}[h_m] = \frac{(c^2+r^2)^2}{r^2},
\hspace{2cm}
\lim_{z_2\to z_1+c\pi}[h_m] = \frac{c^2\pi^2(c^2+r^2)}{4r^2},
\]
which are finite nonzero for positive values of $c$ and $r$. For the last factor, fix $z_1=0$. Then finding when the factor vanishes is equivalent to finding when $\sin(z_2/c)$ and $-z_2 c/r^2$ intersect. There are values for which this happens, and the other factors in $h_m$ are all finite at these values, so $\inf_{z\in \R}[h_i(z)]=0$. Visual confirmation is given by the cases below.
Hence this is not the best approach to calculate the conditioning number of a curve. The next attempt will be to calculate the actual pairwise conditioning number, rather than trying to bound it from below.
\[
\tau_{p,q} = \sup\{\epsilon \ :\ N^\epsilon_pM\cup N^\epsilon_qM\text{ embeds in }\R^n\}.
\]
The condition on $\epsilon$ is the same as saying $i(N^\epsilon_pM)\cap i(N^\epsilon_qM) = \emptyset$, where $i$ is induced by the embedding of $M$. It is immediate that $\tau = \inf_{p,q}\{\tau_{p,q}\}$, so we will try to find $\tau_{p,q}$ first. Recall that a helix of radius $r$ and vertical period $2\pi c$ is a 1-dimensional manifold
embedded in $\R^3$ as the zero locus of
\[
f(x,y,z) = x-r\cos(z/c),
\hspace{2cm}
g(x,y,z) = y-r\sin(z/c).
\]
We first find the normal plane at two arbitrary points $p_1,p_2$ on the helix, then their intersection (which is a line), and then the distance from $p_1$ and $p_2$ to that line. The smallest of these two distances bounds $\tau_{p_1,p_2}$ from below (and the bound is achieved on pairs of points defining the medial axis). Then take the infimum of this value over all points on the helix. However, this excludes the case when the normal planes are parallel (for instance when the two points have the same $x$- and $y$-values).
Moreover, even just calculating the infimum for points whose normal planes are not parallel yields a result of zero. We describe the process nonetheless. For the first step, we need the equations of the normal planes. Let
\[
D^f = \begin{bmatrix} 1 & 0 & r\sin(z/c)/c \end{bmatrix},
\hspace{2cm}
D^g = \begin{bmatrix} 0 & 1 & -r\cos(z/c)/c \end{bmatrix}.
\]
be the Jacobians of $f$ and $g$. The points $p_1$, $p_2$ are completely described by the $z$-coordinate, so we have two values $z_1$, $z_2$ for $p_1$, $p_2$, respectively. The normal plane at $p_i$ is the zero locus of
\[
\det\begin{bmatrix}
x-r\cos(z_i/c) & y-r\sin(z_i/c) & z-z_i \\ 1 & 0 & r\sin(z_i/c)/c \\ 0 & 1 & -r\cos(z_i/c)/c
\end{bmatrix} = z - z_i - \frac{xr}c \sin(z_i/c) + \frac{yr}c\cos(z_i/c).
\]
We have two equations and three unknowns, so one independent variable. Solving for $x$ and $y$ gives us
\[
x = \frac{(z-z_1)\cos(z_2/c) - (z-z_2)\cos(z_1/c)}{r\sin(\frac{z_1-z_2}c)/c},
\hspace{1cm}
y = \frac{(z-z_1)\sin(z_2/c) - (z-z_2)\sin(z_1/c)}{r\sin(\frac{z_1-z_2}c)/c}.
\]
These are functions of $z$, giving us two new functions
\[
h_i(z) = (x(z)-r\cos(z_i/c))^2+(y(z)-r\sin(z_i/c))^2 + (z-z_i)^2,
\]
for $i=1,2$, which, when minimized, give a lower bound for the pairwise conditioning number of $p_1$ and $p_2$. Indeed, by slowly increasing the $\epsilon$ until the $\epsilon$-normal planes at $p_1$ and $p_2$ intersect, the first point of intersection will happen on the intersection $N_{p_1}M\cap N_{p_2}M$. Hence finding the shortest distance from $p_1$ and $p_2$ to this line gives a definite lower bound. The functions $h_i$ are quadratic in $z$, and we know the function $az^2+bz+c$, for $a>0$, has minimum at $-b/2a$. The values of $h_1$ and $h_2$ at their minima are the same and equal to
\[
h_m := h_i\left(\frac{-b}{2a}\right) = \frac{2(c^2+r^2)\cos^2\left(\frac{z1-z2}{2c}\right)\left(r^2+c(z_1-z_2)\csc\left(\frac{z_1-z_2}c\right)\right)^2}{2c^2r^2+r^4+r^4\cos\left(\frac{z_1-z_2}c\right)}.
\]
A natural limit of $h_m$ to consider is $z_2\to z_1$. If any of the factors in the numerator are zero, we also get a minimum, so another limit to look for is $z_2\to cz_1$, which makes the cosine factor zero. These are
\[
\lim_{z_2\to z_1}[h_m] = \frac{(c^2+r^2)^2}{r^2},
\hspace{2cm}
\lim_{z_2\to z_1+c\pi}[h_m] = \frac{c^2\pi^2(c^2+r^2)}{4r^2},
\]
which are finite nonzero for positive values of $c$ and $r$. For the last factor, fix $z_1=0$. Then finding when the factor vanishes is equivalent to finding when $\sin(z_2/c)$ and $-z_2 c/r^2$ intersect. There are values for which this happens, and the other factors in $h_m$ are all finite at these values, so $\inf_{z\in \R}[h_i(z)]=0$. Visual confirmation is given by the cases below.
Hence this is not the best approach to calculate the conditioning number of a curve. The next attempt will be to calculate the actual pairwise conditioning number, rather than trying to bound it from below.
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