## Archive for December, 2008

### A Trickier Curve

December 31, 2008

Yesterday’s parametric curve was a warm up for another I wanted to draw. I’ve played with it a little bit, in the same spirit as yesterday’s, but haven’t gotten an answer yet. It looks sufficiently more complicated, so I thought I’d just go ahead and post my work so far up here, and see if anybody else was interested. Here’s a picture, explanation follows:

The curve I want is the one in the upper left. I’ve decided to base this on the graph of cosine again, so I am thinking

$\begin{array}{rcl}x(t) &=& t+o_x(t)\\ y(t) &=& \cos(t)+o_y(t) \end{array}$

The two other graphs in the picture, on the right, are approximations to $o_y(t)$ (on top) and $o_x(t)$ (on the bottom).

While $o_y(t)$ scares me a bit, I’ve got an idea for finding a formula for $o_x(t)$. Near multiples of $\pi$, the curve looks vaguely like a sine curve with small amplitude and period $\pi/6$ (so, $c_1\cdot \sin(12t)$). Near odd multiples of $\pi/2$, the curve looks more like the curve from yesterday, $c_2\cdot \sin(2t)$. So we want to sort of mix these curves, more of the first near $\pi$-s, and more of the second near $\pi/2$-s.

A decent way to mix two functions, $f_1(t),f_2(t)$ is to use what I’ll call a mixing function $\mu(t)$ (probably other people have a name for it, and probably it’s something I should know, but whatever). The only requirement I have for a function to be a mixing function is that it takes values between 0 and 1 (and continuity is preferred). I’ll mix the two functions as $f(t)=\mu(t)f_1(t)+(1-\mu(t))f_2(t)$. Notice that when $\mu(t)$ is near 1, the mixed function is near $f_1(t)$, and when $\mu(t)$ is closer to 0, the mixed function is near $f_2(t)$.

It’s not too hard to determine that to mix my two sine curves like I want, my $\mu$ will be based on $\cos(2t)$, namely $\mu(t)=(1/2)(1+\cos(2t))$. You can make the mixing ‘quicker’ (more time near each curve, and quick transitions to the other function) via $\mu_p(t)=(1/2)(1+\cos^{p}(2t))$ for small positive values of $p$ (I’m thinking $p=1/n$ for some positive odd integer $n$). To get some idea how this looks, here’s a graph mixing $(.2)\sin(12t)$ with $(1)\sin(2t)$, using $\mu_1$ (the basic $\mu$, fooplot doesn’t seem to want to take odd roots of negative values):

I think that by modifying the various parameters ($c_1,c_2,p$), one could get a pretty decent function for the $o_x(t)$ I was looking for. Like I said, though, $o_y(t)$ looks a bit harder. Perhaps you’ve got a suggestion? I’m tempted to believe you could base it off of $-\cos(t)$… It might almost be a mixing of $-\cos(t)$ with $(1/2)\cos(t)$ (or various other amplitudes on both).

I feel like maybe I should mention why I was thinking these curves, and why I want to draw that parametric curve. When I was playing with my quadric polynomials a while ago, I wanted to draw animations, but I had two parameters floating around. So I thought that if I made a parametric curve that occupied ‘most’ of some square region in the plane, then I could use just one parameter, and make a nice animation. Probably I’m not explaining that too well, but it also isn’t entirely worth it. My goal was to draw a curve that occupied ‘lots’ of the region $-1\leq y\leq 1$ in the plane, and the curve I came up with is the one I’ve been trying to draw in this post.

### A Parametric Curve

December 30, 2008

Today I decided to find parametric equations to graph a curve resembling:

By putting some axes in the same picture, I decided it looked pretty similar to the graph for cosine (shown below in blue):

One may think of $y=\cos(x)$ as a parametric curve given by the equations $x(t)=t, y(t)=\cos(t)$. I want to use this as my base, and modify it just slightly to get the above curve. So I begin by assuming the curve I want has $y(t)=\cos(t)$. Next, I have to think about $x(t)$. I know that if $x(t)=t$, then I’ll get the graph of cosine, so I need to modify $x(t)$ slightly. Perhaps I could make $x(t)=t+o(t)$, thinking of $o(t)$ as some offset function.

What does my offset function look like? Well, comparing the graph I want with the graph for cosine, I see that I want my offset function to be 0 at $t=0,\pi/2,\pi,3\pi/2$, and $2\pi$. From $t=0$ to $t=\pi/2$, the curve I want is to the right of cosine, so the $x$-offset should be positive. Then from $t=\pi/2$ to $t=\pi$, my offset is negative, and then positive and negative again on the intervals $(\pi,3\pi/2)$ and $(3\pi/2,2\pi)$. If I sketch a graph, it ends up looking something like:

That looks very much like $o(t)=\sin(2t)$. Of course, I’m not sure how the amplitude of the offset needs to work out yet, so I’ll work with a family of curves $o(t)=c\cdot \sin(2t)$, and thus the family of parametric equations

$\begin{array}{rcl}x(t)&=&t+c\cdot \sin(2t)\\ y(t)&=&\cos(t)\end{array}$

By playing around with different values of $c$ you get a nice collection of curves:

Here, the black curve is $c=1$. Then they go red, blue, green, orange, increasing $c$ by .5 each time. So in the end, the orange curve has $c=3$.

Notice that sometime between $c=2$ (blue) and $c=2.5$ (green), the curve starts having self intersections. So the obvious question is, what value of $c$ is the smallest that gives self intersection? Judging by the pictures, this will also cause a shared vertical tangent at each intersection point, which will have $x$-coordinate $\pi$.

Well, ok, vertical tangents occur when $x'(t)=0$ (and $y'(t)\neq 0$), so find $x'(t)=1+2c\cdot \cos(2t)$ and solve to find $t=\frac{1}{2}\arccos(-1/(2c))$. Call this value $t_0$. We want $x(t_0)=\pi$, so we have to solve $\frac{1}{2}\arccos(-1/(2c))+c\cdot \sin(\arccos(-1/(2c)))=\pi$. Not something I know how to do. Sure, you can simplify $\sin(\arccos(A))$ using known formulas, but it doesn’t help much. Another idea is to use a double-angle identity to write $x'(t)=1+2c(1-2\sin^2(t))$, with a zero at $t=\arcsin\left(\sqrt{\frac{2c+1}{4c}}\right)$. This is mildly optimistic, because we now have $\sin$ mixing with $\arcsin$, but it still doesn’t seem to get us very far. I messed about for a while on Maple and came up with a decimal value around 2.315643114. I’ve not thought of another way to do this, or how to solve those equations, perhaps somebody’s got an idea?

I mentioned that I was thinking about these things on twitter, and @samjshah replied:

i think the exact answer for c is the solution to this eqn: .5arccos(-1/(2c))+.5sqrt(4c^2+1)=pi. great problem!

Looks like we were thinking along the same lines (I’ve @replied him, to see if he’d like to leave a comment below).

So anyway, that’s about as far as I’ve taken it, and probably about as far as I plan to go. For grins, I also plotted some negative values for $c$:

The colors correspond to coefficients as in the previous picture, it’s just the negative value of the constants (so, e.g., black corresponds to $c=-1$). Looks like if we keep going (larger negative values) we’ll start getting more self intersections, so again you might ask where. Seems like it should be a pretty similar exercise (I didn’t assume $c>0$ in my work above – at least, not on purpose).

In case you were curious, I made all of the above pictures using fooplot.com, which I’ve found to be pretty handy. It’s also fun to make animations in Maple, using something along the lines of (after a with(plots)):

display(seq(plot([t+(N/10)*sin(2*t),cos(t),t=0..2*Pi]),
N=-30..30),
insequence=true);


### Meh

December 15, 2008

Progress has been, as usual, slow. And I haven’t been updating here. My advisor’s last advice was to try to avoid the category $\mathscr{L}_M$ I mentioned previously (category of locally affine partitions on $M=m\times \mathbb{R}^n$). Ideally I can use something like the $\mathscr{C}_M$ category and just find appropriate reductions. I should be looking to deformation retract $\mathscr{C}_M$ down to a smaller category, and also deformation retract $\mathscr{J}_M$ (which I mostly know has the right homotopy type, in the holim) down to the same category. Or something like that.

So I’ve had two sorts of thought strands going. This is not unexpected, as my overall project is really just the blending of two known cases (some of my advisor’s work). One thing I’ve been thinking about is when $M=m\times \mathbb{R}^0$ is just a finite set. And I’ve been thinking about a subcategory $\mathscr{S}_M$ of finite sets with partitions, as opposed to the full category $\mathscr{C}_M$ of abstract completely locally affine partitions (what a mouthful), with non-locally-constant maps to $M$. I think my goal is to deformation retract $\mathscr{S}_M$ down to something along the lines of those pairs $(\lambda,f)$, where $\Lambda \vdash s\xrightarrow{f}M$ and $f$ is injective on the components of $\Lambda$ (or, perhaps even injective overall). To do this, I gotta get the right topology on $\mathscr{S}_M$, which I’m pretty sure I’ve not yet done.

Part of the reason I think I’ve not yet done this is based on my other line of thought, concerning the case $M=1\times \mathbb{R}^n$ (frequently $n=1$ when I’m thinking about it). I’m still thinking about a subcategory of $\mathscr{C}_M$ consisting of just finite sets (with partitions) and non-locally-constant maps to $M$. But I need to set things up so that “points coming together” corresponds to a different partition. That is, when I have two points in the same equivalence class of $\Lambda$ and their image in $M$ lie in the same component (which is easy when $M=1\times \mathbb{R}^n$), then I could make a sequence of functions that are non-locally-constant for this $\Lambda$ and bring the two points together. The limiting point should be considered as a map from a smaller $\Lambda$. Another way to say this is that I should think of points in “thick diagonals” (some, but not all, points are the same, e.g. where $x=y$ in $\mathbb{R}^3$) of $M^n$ as coming from some “smaller partition”.

I’m not saying any of this well. Probably that’s a bad sign. Perhaps I’ll try again tomorrow. Or after I run some ideas by my advisor on Tuesday. Anyway, half the point of this post was to show off the pictures I was looking at today. I was trying to draw my category (or small portions of it) in the case $M=m\times \mathbb{R}^0$, and was, for some reason, pleased with the following picture:

I don’t think I’ll explain too much about it, since I’m not sure how useful it’ll be long-term. But if you’ve thought about set partitions of the set $\underline{4}={1,2,3,4}$, you might recognize the poset (ordered by refinement) in the diagram above. For any partition $\Lambda$, I have written down the possibilities for maps to $M=\underline{2}\times \mathbb{R}^n$ such that the composite with the projection $M\to \underline{2}$ is non-locally-constant on blocks of $\Lambda$. Those underlined in red are, furthermore, injective on blocks.

### More Catchup

December 1, 2008

My last post started laying down some of the framework for what I’ve been thinking about, and I thought I’d continue on with that here.

Recall $M=m\times \mathbb{R}^n$, and that I have a category $\mathscr{C}$ which I think of as a category of “complete locally affine partiitons”. From this I construct the category $\mathscr{C}_M$ consisting of pairs $(\rho,f)$, where $\rho\in \mathscr{C}$ (so $\rho=(\rho,A_{\rho},s,\Lambda)$) and $f:A_{\rho}\rightarrow M$ does not factor through $\rho$ (that is, there are two points that are equivalent in the partition $\rho$, but their images are different in $M$). I’ve also taken up the habit of saying that these maps “disrespect” $\rho$. Out of this category I had a functor $nlc(-,V)$, the non-locally-constant maps to a real vector space $V$. Of course, since this is supposed to a be “topological category” (I have a space of objects, and a space of morphisms…), a functor from the category is really a fiberwise space over the space of objects. Let me denote this space by $\mathscr{N}_{\mathscr{C}_M}$.

I have another category that I didn’t mention yesterday. I denote it by $\mathscr{L}_M$, and it consists of locally affine partitions of $M$. Given two locally affine partitions, $\sigma,\sigma'$ in this category, there is an arrow $\sigma\to\sigma'$ precisely if $\sigma\leq \sigma'$ (that is, $\sigma$ is coarser than $\sigma'$ – if two elements are related by $\sigma'$ then they are related by $\sigma$).

I’m being a little imprecise with the definition of this category for now, because I’m still playing with it and trying to figure out what makes it workable. The idea is that there is a functor $im:\mathscr{C}_M\to \mathscr{L}_M$ given by taking the image. That is, a point in $\mathscr{C}_M$ is a pair $(\rho,f)$, and it makes sense to consider $f(\rho)$, whose affine completion is a locally affine partition of $M$. Out of the category $\mathscr{L}_M$ we have yet another “non-locally-constant” functor, $nlc(-,V)$. Given a locally affine partition $\sigma$, $nlc(\sigma,V)$ will be maps $M\to V$ that disrespect $\sigma$. Of course, I’m again in the situation of topological categories, so really this functor is an object $\mathscr{N}_{\mathscr{L}_M}$ over the space of objects of $\mathscr{L}_M$.

I can now take the pull-back of $\mathscr{N}_{\mathscr{L}_M}$ along $im$, obtaining an object (functor) which I’ll denote $\mathscr{N}_{\mathscr{L}_M}^*$ over $\mathscr{C}_M$. so, if you are counting along at home, that’s two functors over $\mathscr{C}_M$, and one over $\mathscr{L}_M$. With any luck I’ll be able to compare all the associated homotopy limits.

There’s one more category, which I’ve been calling $\mathscr{J}_M$. The letter ‘J’ comes from the fact that this category comes out of some join construction, based on a theorem of Thomason for manipulating (ho)limits (and the general brilliance of my advisor). I’ve got a functor out of it, and I know that the homotopy limit of this guy is the space of embeddings I want. When I get $\mathscr{L}_M$ sorted out, there should be a nice obvious functor $\mathscr{J}_M\to \mathscr{L}_M$ (essentially an inclusion), and that’ll give me the ways to tie together all of my homotopy limits, and tie them all to the space of embeddings. But that’ll have to wait for another day. I was tired when I started this post, and it’s only gotten worse.