I guess this is somewhat blog-like. Note that I’m still working on this page, and trying to bring stuff from my old site over (that is, this content is incomplete).

12 August 2007

Note: in retrospect, you can just do this with the Riemann sphere. I didn’t know this at the time.

Figure 1. Graph of y = 1/(100x).

I’ve been thinking of a method to graph a function over the entire coordinate plane and then mapping it to the surface of a sphere. For mapping all real numbers to a single interval, I realized the inverse tangent function would work. On a side note, does this mean that there are as many real numbers from −π to π as there are from −∞ to ∞?

Then, taking this mapping from infinite to finite intervals, another transformation is needed to map the image onto a sphere. In reality, there is no point to proceed with this mapping since a [−π, π] × [−π, π] square is more informative and can be better represented in 2D media, but it is cool to see something like the inverse function mapped onto a sphere (as represented to the right) since the discontinuities seem to loop around to infinity and back. Although, this may not be the best way to think of functions like this.

Figure 2. Graph of y = 
sin x/5.

The way I chose to do this mapping involved looking at the great circle between the zero and infinite points, which are on opposite ends of a diameter. Then, to plot a single point on the sphere, the transformed x-coordinate is the angle around the great circle with respect to the zero point to construct another circle with this as one point on the circumference, the infinity point being the another point on the circumference, and the center being contained within and perpendicular to the plane of the great circle. Since the slice of a sphere is a circle, this constructed circle is contained on the surface of the sphere. The transformed y-coordinate is now the angle around this constructed circle with respect to the point on the great circle. The point at this angle is the image of the complete transformation from the cartesian plane.

So, let f(x) be a function R2 → R2 that maps all points of the plane to the region [−π, π] × [−π, π]. For instance,

f(x) = (
2tan −1x1
2tan −1x2

Then, let g(x) be a function R2 → R3 that maps the region [−π, π] × [−π, π] to the surface of a sphere centered at the origin with a radius of one. After a bit of computation,

g(x) = 
(cos x1 + 1) ⋅ cos x2 + cos x1 − 1
sin x1 ⋅ cos x2 + sin x1
sin x2 ⋅ (2cos x1 + 2)1/2

So, the mapping of the coordinate plane onto a sphere is a function R2 → R3 defined by h(x) = g(f(x)).

Figure 3. Graph of y = x.

For the purposes of displaying a graphic of these plots, a linear transformation can be used from R3 to R2 that projects vectors onto a plane while scaling to get a more pleasing image. It results in the z-axis pointing up, the x-axis to the left, and the y-axis to the right, maintaining the right hand rule. This was constructed by an ad-hoc methad.

Tx = (
− 1/2 − 1/61

This transformation is then applied to h(x) to yield something similar to one of the plots on this page.

This transformation currently distorts functions quite a bit near infinity since it is asymmetric with respect to the images of the x- and y-axis. A good example of the problems inherent in the transformation can be seen in an examination of the function f(x) = x which one would expect to be a great circle with an equal angle between the two axis. But, it gets a strange little curve near infinity, as Figure 3 illustrates.


While doodling, I found that a closed loop, under the right circumstances, can be colored using only two colors. In this case I used white and gray. Here are definitions and the theorem. The definitions are my own and may conflict with actual established mathematical language.

Figure 4. Example of theorem
Simple intersection
a non-tangential intersection of exactly two curve segments.
A set of regions is colored only if no two adjacent regions have the same color.
The regions enclosed by a smooth, simply intersecting, closed curve can be colored with at most two colors.

I have an idea for how to prove this by using a set of reduction rules to turn the curve into a bunch of circles (look at Figure 4 and pretend each gray blob can be broken from each other gray blob at the intersections. The maximum number of colors that you need for a bunch of separate circles is two: the background, and the color of the circles).

5 July 2007?
I was watching the fireworks on the 4th of July and I was timing how long it took for the sound to reach me. I estimated it was roughly 2.5 seconds. Then, I wondered what the show would be like if it were recorded and synched with the sound track set back 2.5 seconds so the explosion and the sound coincided.