Wrapping Up Infinite Planes

Note: This is sort of a “part 2” to this post, which is the two-dimensional analogue to this (mainly) three-dimensional post.

It’s fairly early/late (4:39AM) and I’m pretty sleepy but I wanted to try and get this post in, so I apologize in advance if it seems “more math-y and less descriptive.”sphere.png

As an expansion to the previous post, the same concept can be applied to the sphere x^2+y^2+z^2=1 and \mathbb{R}^2. And similar to before, the top of our sphere is punctured, i.e. the point (0,0,1).

We now look for a function f that allows us to input a point (x_0, y_0) on our plane and output a point (x_s,y_s,z_s) on our sphere, that is, f: \mathbb{R}^2 \to \mathbb{S}^2, where \mathbb{S}^2 = \{ x \in \mathbb{R}^3 | x^2+y^2+z^2=1 \}. The line that will assist us with this assignment is now a vector, as would be required, so here’s our new system:

\begin{cases} x^2+y^2+z^2=1 \\ \vec{r}(t)=\left< 0,0,1 \right> + t \left< x_0,y_0,-1 \right> \ \ \ 0 \leq t \leq 1 \end{cases}

Another, more convenient way of writing this system can be done by noting that \vec{r}(t) can simply be rewritten (by adding the vectors in its definition) as

\vec{r}(t) = \begin{cases} x(t) = x_0 t \\ y(t)=y_0t \\ z(t) = 1-t \end{cases} = \ \ \ \begin{cases} t=\frac{x}{x_0} \\ t=\frac{y}{y_0} \\ t = 1-z   \end{cases}

thus giving us

\begin{cases} x^2+y^2+z^2=1 \\ \frac{x}{x_0}=\frac{y}{y_0}=1-z \end{cases}

by equating the the parameter in each component. From here we just need to take turns solving for x, y, and z in terms of x_0 and y_0 and we should be all set!

Solving for x:
\displaystyle{ x^2 + \left( \frac{xy_0}{x_0} \right) ^2 + \left( 1-\frac{x}{x_0} \right) ^2 = 1}
\displaystyle{ \left( \frac{{x_0}^2+{y_0}^2+1}{{x_0}^2} \right)x^2 - \frac{2}{x_0} x = 0}
\displaystyle{ x=\frac{\frac{2}{x_0} \pm \sqrt{\frac{4}{{x_0}^2}}}{2 \left( \frac{{x_0}^2+{y_0}^2+1}{{x_0}^2} \right)}      }

\displaystyle{ \implies x = \frac{2x_0}{{x_0}^2+{y_0}^2+1}}

Solving for y:
\displaystyle{ \left( \frac{yx_0}{y_0} \right) ^2 +y^2 + \left( 1-\frac{y}{y_0} \right) ^2 = 1}
\displaystyle{ \implies y = \frac{2y_0}{{x_0}^2+{y_0}^2+1} }

Note, for x and y, the radical was chosen to be positive or else both would’ve reduced to to being 0 for all values of x_0 and y_0, which is not always the case, for instance at (x_0, y_0) = (1,0), that should be mapped to the sphere at the point (x_s, y_s, z_s) = (1,0,0) on the sphere, and similarly (0,1) \mapsto (0,1,0), thus indicating we take the positive root.

Solving for z:
Let’s let u ={x_0}^2+{y_0}^2.
\displaystyle{ z = \frac{2u \pm \sqrt{4u^2-4(u+1)(u-1)}}{2(u+1)}}
\displaystyle{ z = \frac{u \pm 1}{u + 1}}
Now taking the numerator that doesn’t assign z to the punctured point…
\displaystyle{ \implies z= \frac{{x_0}^2+{y_0}^2-1}{{x_0}^2+{y_0}^2+1}  }

Our Function:
Using the above information, our function f: \mathbb{R}^2 \to \mathbb{S}^2 is defined

\displaystyle{ f(x_0, y_0) = \left< \frac{2x_0}{{x_0}^2+{y_0}^2+1}, \frac{2y_0}{{x_0}^2+{y_0}^2+1},\frac{{x_0}^2+{y_0}^2-1}{{x_0}^2+{y_0}^2+1}  \right> }

and there we have it!

Now, this function is actually bijective as well. Go ahead and find the explicit inverse! That is, if I give you a (x_s, y_s, z_s) and you tell me what (x_0, y_0) was assigned there.

On another note, doesn’t it seem like the function about really resembles the one from our previous post? Our other function f: \mathbb{R} \to \mathbb{S} was defined to be f(x_0)=\displaystyle{ \left< \frac{2x_0}{{x_0}^2+1}, \frac{{x_0}^2-1}{{x_0}^2+1} \right>}. You know, I dare be so bold as to assert the following.

The function f: \mathbb{R}^n \to \mathbb{S}^n where \mathbb{S}^n = \{ x \in \mathbb{R}^{n+1} : ||x|| = 1 \} defined by

\displaystyle{ f(x_1, \dots, x_n) =  \left< \frac{2x_1}{\sum_{i=1}^n x_i^2\ + 1}, \frac{2x_2}{\sum_{i=1}^n x_i^2\ + 1}, \dots, \frac{2x_{n-1}}{\sum_{i=1}^n x_i^2\ + 1}, \frac{\sum_{i=1}^n x_i^2\ - 1}{\sum_{i=1}^n x_i^2\ + 1} \right> }

is bijective.

Proof: Too tired, but I’ll leave it up to the reader to prove/disprove.

The set \mathbb{R}^n is isomorphic to \mathbb{S}^n.

Note, ||(x_1, \dots, x_n) || = \sqrt{{x_1}^2 + \dots + {x_n}^2}.

Another way of proving the above conjecture is finding the horrendous function’s inverse (the answer is called the stereographic projection, but there’s no fun in googling first!).

Also, the conjecture implies that \mathbb{R}^n and \mathbb{S}^n, which is really worth giving it some thought if this is your first time hearing this. That radius 1 circle from last post? It has as many points as the real number line. The sphere of radius 1 in this post? Has as many points as the infinite two-dimensional plane of real numbers. Just let that set in.

Has the existential crisis kicked in yet?

On a very final note, if you thought that this and the preceding post were interesting I highly recommend looking into stereographic projections. Also, if you want a sort of “interactive” program to mess around with spherical geometry and projections onto a plane, take a look at this java program called Sphaerica. Here’s some screenshots from it:

Well, that’s all I got for today. Probably will be a good few days until my next post so I can mainly focus on my studies. See you then!


Wrapping the Real Numbers Around a Circle… of Radius 1

I’d like to thank my amazing Calculus III instructor, Professor Wolfson, for showing me this. Undoubtedly will always be my favorite professor and I only wish to someday be a quarter as knowledgeable as him. Anyhow! Let’s jump right in.

What if I told you that you can wrap the entire real number line, every point on it, around a circle of radius 1? Go ahead, choose whether I’m a loony or not now before continuing to read.

Did you choose? Alright. Now look at the animated graph below.

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Circle has radius 1, line is in the form y=mx+1 with -10<m<10.

I’m going to try to motivate you to sort of “prove” this in your head before I get into the mathematics behind it. While looking at the gif, just note the following:

  1. Consider the very top point of the circle, the point (0,1), as undefined.
  2. Note that as the slope changes, where the line intersects the circle and the x-axis changes as well. In a single snapshot of the gif, let’s call the point it intersects on the x-axis as (x_0, 0) or just x_0 for short. Call the point it intersects on the circle (x_c,y_c)
  3. Now try to imagine this: in that random snapshot of the gif (where the line is held still) map (“assign”) the point x_0 to (x_c,y_c). Now go to the next snapshot and do it again, and again, …

Now let the slope instead of being anything in the range (-10, 10), be anything in ( - \infty, \infty). What follows is every single point on the real number line (x-axis) gets assigned to a point on the circle! With room to spare even, that point on the top of the circle never got assigned anywhere after all.

Let’s get down and dirty in some math now.

Err, I’m sorry, I won’t say that again. Sounded a lot cooler in my head.

Finding the Function:
We can actually find the exact function f that tells us where our point x_0 goes on the circle. It’ll be a function mapping one number to two, or more explicitly, f: \mathbb{R} \to \mathbb{S}, where \mathbb{S} = \{ x \in \mathbb{R}^2 | x^2+y^2 =1 \}.

To get started we need to know what we have to work with. Without loss of generality let’s let the radius be 1 (that is, you can make the radius whatever you please, I just wanted to make it look pretty in the end). What we have is

\begin{cases} x^2+y^2=1 \\ y=mx+1 \text{ where } m = -\frac{1}{x_0} \end{cases}

which is actually all we need. Let’s leave the m in there for now though to avoid having to deal with messy equations, and just remember that our circle doesn’t include the point (0,1).

From here it’s a matter of some plug and chug!

\displaystyle{ x=\frac{-2m \pm \sqrt{4m^2}}{2(1+m^2)}}
\implies x=\displaystyle{\frac{-2m}{1+m^2}}

Great! Now we complete the same procedure for the other coordinate.

\displaystyle{\left(\frac{y-1}{m} \right) ^2 + y^2 = 1}
y=\displaystyle{\frac{2 \pm \sqrt{4-4(1+m^2)(1-m^2)}}{2(1+m^2)}}
y= \displaystyle{\frac{1 \pm m^2}{1+m^2}}
Here we must choose 1-m^2 for the numerator because otherwise things would simplify to y=1, and remember, (0,1) wasn’t defined on our circle.
\displaystyle{ \implies y=\frac{1-m^2}{1+m^2}}

The Function:
All this, in summary, provides the following function:
f(x_0)=\displaystyle{ \left< \frac{-2m}{1+m^2}, \frac{1-m^2}{1+m^2} \right>}
which, after plugging in the fact that m=-\frac{1}{x_0} provides…

f(x_0)=\displaystyle{ \left< \frac{2x_0}{{x_0}^2+1}, \frac{{x_0}^2-1}{{x_0}^2+1} \right>}

And there we have it! Any point x_0 on the real number line can be mapped to the point f(x_0) on our circle of radius 1! Go ahead and try a few points and mess around if you want. Let’s end on some final notes regarding this:

  • It’s pretty cool looking at how “dense” the circle is packed, for instance, the bottom half of the circle is mapped only points from [0,1] but what about perhaps the upper-left tiny little 1/1000 segment of the circle? Oh dear…
  • Thing about the difference in the arc-length of a circle over its entire domain [0, 2\pi] versus the arclength of x=0 over entires domain (-\infty, \infty). What in the world is going on here?
  • Explicitly calculate the inverse of f(x_0), that is, if I gave you a point (x_c, y_c) on the circle, can you tell me what x_0 it was assigned?
    • By finding the explicit inverse function (showing it exists), we showed that the circle and the real numbers are isomorphic (have the “same size” set of points!!).

Edit: Part 2 of this is up!