## Thor’s Day Morning Mathematical Musings

1. Can you completely mix a mug of coffee, such that, at every point inside the mug, the coffee at that point is different after stirring from before stirring? Go get a cup of joe (or tea), stir it, and see what you think.
Answer: no. There will always be at least one point that is the same after the liquid has settled, no matter how vigorously you stir it. It is mathematically impossible for there to be no such points inside the mug.
2. Do there exist on the surface of the Earth, at any given time, two antipodal points that have exactly the same surface temperature?

Answer: yes. What about two antipodal points that have exactly the same barometric pressure? Also yes. Two antipodal points that have exactly the same surface temperature and exactly the same barometric pressure? Yet again, yes. This is mathematically inescapable.‌

At any time there exists a continuous curve on the Earth’s surface on which every point has an antipodal point that also lies on the curve and that has the same temperature. There is a different continuous curve on which antipodal points have the same pressure. And the two curves must intersect, since both encircle the globe, each separating it into two pieces. So that means there must be, at any time, at least one pair of antipodal points somewhere on the surface of the Earth that have the same temperature and the same pressure.

You’ve probably surmised by now—you drank that cup of joe, right?—that this is true not just for temperature and pressure but for any two continuously variable parameters (such as temperature, pressure, humidity, wind speed, solar and terrestrial radiation, cloud ceiling, particulate density, atmospheric composition, and so on). You would be correct.
3. Think of a multi-digit positive integer. Any such number will do—for example, $76.$ Now add up its digits and subtract that sum from the original number. $76\,- (7+6) = 63.$ Now apply this algorithm to the new number: $63\,- (6+3) = 54.$ Keep doing this until the resulting number has shrunk to just one digit. $54\,- (5+4) = 45$, $\dots, 18\,- (1+8) = 9.$

Ta da! (Yes, really.) No matter your starting number (as long as it has more than one digit), you will always end up at $9$.

Here is a quick and dirty python program that performs this task for any positive integer, returning the end result (which had better be nine!) and the number of iterations it took to get there:

def digi9(n):
count = 0
while True:
k = sum(list(map(int,','.join(str(n)).split(','))))
m = n - k
if len(str(m)) == 1:
return m, count+1
n = m
count += 1

Let’s consider an example:

>>> digi9(72459075)
(9,2191634)

Starting with the randomly chosen number $72,459,075$, over two million iterations later we indeed end at $\dots, 27\,-(2+7) = 18,$ $18\,- (1+8) = 9.$

How are the answers to these little puzzles so? Welcome to the world of fixed point theorems! In mathematics, a fixed point is a member of a set such that an operation on the set at that point maps back to the point. The set can be anything—the set of integers, a Euclidean line, surface, or volume, etc. This concept has wide application and profound consequences in many branches of mathematics. The above puzzles are examples of fixed points in their respective sets. Put that in your mug and stir it!

Now go get some more coffee.

#### Show Me!

Suppose we have a function $f(x)$ such that $f(x) \in [a,b]~~\forall~x \in [a,b]$. That is, the function maps back to its domain. Then $f(x)$ has a fixed point $f(c) = c$ somewhere in the closed interval $a \le c \le b$.

Why? Well, it must be true that

$$f(a) \ge a~~~ \mathrm{and} ~~~f(b) \le b \label{condition}$$

The intermediate value theorem says that if a function $f(x)$ is continuous on a closed interval $[a,b]$, then, for a given $c$ such that $f(a) \le c \le f(b)$, there must exist at least one value $x_0 \in [a,b]$ such that $f(x_0) = c.$

Since the range of our function is restricted to its domain, $f([a,b]) \in [a,b]$, we have from eq. \eqref{condition} that $f(a)-a \ge 0$ and $f(b)-b \le 0.$ If we define $g(x) \equiv f(x)-x$, this is $g(a) \ge 0 \ge g(b).$ By the intermediate value theorem there must then exist a value $c \in [a,b]$ such that $g(c) = 0$. Hence, there must exist at least one fixed point, $f(c) = c.$

This—or, rather, its generalization to any Euclidean space—is essentially a statement of the Brouwer fixed point theorem:

Every continuous function from a closed ball of a Euclidean space into itself has a fixed point.

Legend has it that Brouwer was lead to his theorem by pondering the surface of a cup of coffee upon stirring in a lump of sugar. (That someone would debase a good cup of coffee with sugar is a wholly different issue.)

Vsauce has an interesting video about fixed points, from which I stole the three examples above:

## Prolegomenon

You recognize as a youngster that science, and music, and literature and writing—creative wonders—draw you along comfortable invisible force lines. But not opera. Overbearing, embarrassing falsetto vibrato is just wrong. As your joints grow creaky and more of your pate warms to the Sun, you know that this is a misperception. You stumble upon more of these, as you notice yourself more often assigning past vigorous feats of physical prowess to the unimportant pursuits of the unimportant young. You ponder these, your various misperceptions. And your misperceptions of misperceptions. Recursion tickles you.

$$\dfrac{\mathrm{d}^2\overrightarrow{r}}{\mathrm{d}\theta^2}+2\widehat{z}\times\dfrac{\mathrm{d}\overrightarrow{r}}{\mathrm{d}\theta}+{\left(\widehat{z}\cdot\overrightarrow{r}\right)}\widehat{z}=\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\overrightarrow{\nabla}\mathrm{\Omega}$$

You realize in the shower one day that your—and others’—universal cognitive foibles smacking into observable reality are an irresistible rabbit hole, wondrously vast and an endless source of material to contemplate. Like a particle in the three-body problem of celestial mechanics, your orbit is a tangled meandering, variously lured into the sphere of influence of first one and then the other of those two massive attractors, science and the creative urge. This resonates, and you realize a re-appreciation of past love.

$$\mathrm{\Omega}=\frac{1}{2}r^{2}+U=\frac{1}{2}r^{2}+\frac{{1-\mathrm{\mu}}}{r_{1}}+\frac{\mathrm{\mu}}{r_{2}}$$

Thus: what shall you write? Unuseful question. The world is big. Where shall you intend your aim? Better. Get thee to the shower!, your ever-reliable Delphic font of nearly every good idea.§ You love nature, and science—especially astronomy and math—and the scientific way of thinking, which come to you with joy and not pain. (This cannot be weird, surely—friends’ and society’s protestations notwithstanding.) The chasm awaits.

$$r_{1}=\sqrt{{{\left(x+\mathrm{\mu}\right)}^{2}+y^{2}+z^{2}}}\hspace{2.222222em}r_{2}=\sqrt{{{\left(x-1+\mathrm{\mu}\right)}^{2}+y^{2}+z^{2}}}$$

On a whim you schlep to a National Association of Science Writers conference, where you are isolated and small, sole introvert amidst a mind-bruising cacophony. Drilling through your crushing discomfort, you meet Roy Peter Clark’s Writing Tools: 50 Essential Strategies for Every Writer (you buy three copies), you hear Jonathan Coulton sing his wistful nerd anthem, “Code Monkey” (you buy three CDs), and a merciful soul tells you to read Lewis Thomas’s classic medley of essays, The Lives of a Cell: Notes of a Biology Watcher (why is there no Kindle version?). This is it. A trigger, an unlatching: your dormant writing compulsion awakens.

Astronomy with math. True stories, precisely told. A worthwhile target.

$$v^{2}-\frac{{2\mathrm{\Omega}}}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}+z^{2}+C+2\int\frac{{e_{p}\mathrm{sin}\mathrm{\theta}}}{{\left(1+e_{p}\mathrm{cos}\mathrm{\theta}\right)}^{2}}\mathrm{\Omega}\hspace{0.222222em}d\mathrm{\theta}=0$$

Halfway through college, you end the pleasant agony and decide astronomy over music. Seemingly by crazy random utterly naive inevitability, you become a professional astronomer. As your mop grows thinner and your knuckles grow larger, you realize the apparent randomicity is a misperception.

The equations, if you are wondering, tell how a massless particle moves in the combined gravitational fields of two massive objects in orbit about each other.¤ Think, for example, Sun–Jupiter–spacecraft. In astronomy, we call this the restricted three-body problem. It is astonishingly complex.

§ Perhaps only Death is a greater surety—though, surely, only by a little.

¤ For completeness:

$$\mathrm{\mu}=\frac{m_{2}}{{m_{1}+m_{2}}},\hspace{2.2em}r=\sqrt{x^2+y^2+z^2}$$

and

$$\begin{array}{rcl}\overrightarrow{\nabla}\mathrm{\Omega}&=&\left[\begin{array}{l}x-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}\left(x+\mathrm{\mu}\right)-\dfrac{\mu}{r_{2}^{3}}\left(x-1+\mathrm{\mu}\right)\\\\y\left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mu}{r_{2}^{3}}\right)\\\\z\left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mu}{r_{2}^{3}}\right)\end{array}\right]\\\\&=&\left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mu}{r_{2}^{3}}\right)\overrightarrow{r}-\mathrm{\mu}\left(1-\mathrm{\mu}\right)\left(\dfrac{1}{r_{1}^{3}}-\dfrac{1}{r_{2}^{3}}\right)\widehat{x}\end{array}$$

## math test

Here’s how to get MathJax up and running for your blog: part I, part II. The three tests below are text lifted from elsewhere.

### Test 1

Consider first what we shall call the direct geometry case, in which we use only the zenith angle $z$ and bypass the geocentric angle $\theta$. The length of side $\overline{CM}$ follows from the right triangle $\widehat{CMP}$:

$$$$\begin{array}[b]{ccl}\left(R+H\right)^{2} & = & \left(D\sin z\right)^{2}+\left(R+h+D\cos z\right)^{2}\\ \\ & = & D^{2}+\left(R+h\right)^{2}+2\left(R+h\right)D\cos z\end{array}\label{eq:R+H-test}$$$$

or

$$D^{2}+2\left(R+h\right)D\cos z-\left[\left(R+H\right)^{2}-\left(R+h\right)^{2}\right]=0\label{eq:D eqn-test}$$

with solution

$$\begin{array}[b]{ccl}D & = & -\left(R+h\right)\cos z\pm\sqrt{\left(R+h\right)^{2}\cos^{2}z+\left[\left(R+H\right)^{2}-\left(R+h\right)^{2}\right]}\\ \\& = & \left(R+h\right)\left(\sqrt{\cos^{2}z+\dfrac{\left(R+H\right)^{2}-\left(R+h\right)^{2}}{\left(R+h\right)^{2}}}-\cos z\right)\end{array}\label{eq:D soln quadratic ugly-test}$$

where the geometry of the problem requires the positive root. For convenience, define

$$\epsilon\equiv\dfrac{H}{R}\quad\mathrm{and}\quad\xi\equiv\dfrac{h}{R}\label{eq:eps and xsi defs-test}$$

Then we can write eq. \eqref{eq:D soln quadratic ugly-test} as

$$D=\left(R+h\right)\left(\sqrt{\cos^{2}z+\left(\dfrac{1+\epsilon}{1+\xi}\right)^{2}-1}-\cos z\right)\label{eq:D soln quadratic-test}$$

Eq. \eqref{eq:D soln quadratic-test} has the disadvantage of subtraction of two nearly equal numbers.

### Test 2

We would like to know what is the radius $\bar{r}$ of the center of mass

of a grid cell of inner radius $r_{1}$ and outer radius $r_{2}$. In polar coordinates $\left(r,\theta\right)$ an infinitesimal area element is $dA=r\,dr\,d\theta$, so

$$\bar{r}=\frac{1}{\Delta A}\intop_{0}^{\Delta\theta}\intop_{r_{1}}^{r_{2}}r\,dA=\frac{1}{\Delta A}\intop_{0}^{\Delta\theta}\intop_{r_{1}}^{r_{2}}r^{2}dr\,d\theta\label{eq: area-weighted r integral-test}$$

where $\Delta A=\frac{\Delta\theta}{2\pi}\cdot\pi\left(r_{2}^{2}-r_{1}^{2}\right)$.

Thus,

$$\Delta A=\frac{\Delta\theta}{2}\left(r_{2}^{2}-r_{1}^{2}\right)\label{eq: cell area-test}$$

and

$$\bar{r}=\frac{1}{3}\frac{\Delta\theta}{\Delta A}\left(r_{2}^{3}-r_{1}^{3}\right)=\frac{2}{3}\frac{r_{2}^{2}+r_{1}r_{2}+r_{1}^{2}}{r_{1}+r_{2}}\label{eq: area-weighted r-test}$$

[…]

Thus, we have the bootstrapping scheme

$$\begin{array}{rclcrcl}\bar{r}_{0} & = & \dfrac{2}{3\Delta^{2}}\left(r_{2,0}^{3}-r_{1,0}^{3}\right) & & r_{2,0} & = & \sqrt{r_{1,0}^{2}+\Delta^{2}}\\& \vdots & & & & \vdots\\\bar{r}_{k} & = & \dfrac{2}{3\Delta^{2}}\left(r_{2,\,k}^{3}-r_{2,\,k-1}^{3}\right) & & r_{2,\,k} & = & \sqrt{r_{2,\,k-1}^{2}+\Delta^{2}}\\& \vdots & & & & \vdots\\\bar{r}_{N_{r}-1} & = & \dfrac{2}{3\Delta^{2}}\left(r_{2,\,N_{r}-1}^{3}-r_{2,\,N_{r}-2}^{3}\right) & & r_{2,\,N_{r}-1} & = & \sqrt{r_{2,\,N_{r}-2}^{2}+\Delta^{2}}\end{array}\label{eq: bootstrap scheme}$$

where, again, we start with $r_{1,0}=r_{min}$ .

### Test 3

Now, $-\widehat{z}\times{\left(\widehat{z}\times\overrightarrow{r}\right)}=\overrightarrow{r}-{\left(\widehat{z}\cdot\overrightarrow{r}\right)}\widehat{z}$, so

$$\overrightarrow{r}^{\prime\prime}+2\widehat{z}\times\overrightarrow{r}^{\prime}=\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}{\left(\overrightarrow{r}+\overrightarrow{\nabla}U\right)}-{\left(\widehat{z}\cdot\overrightarrow{r}\right)}\widehat{z}\label{}$$

Define a new effective potential

$$\mathrm{\Omega}=\frac{1}{2}r^{2}+U=\frac{1}{2}r^{2}+\frac{{1-\mathrm{\mu}}}{r_{1}}+\frac{\mathrm{\mu}}{r_{2}}\label{EQUATION.5d0b51dc-3a17-4d57-95ed-8e8768257778}$$

where

$$r_{1}=\sqrt{{{\left(x+\mathrm{\mu}\right)}^{2}+y^{2}+z^{2}}}\hspace{2em}r_{2}=\sqrt{{{\left(x-1+\mathrm{\mu}\right)}^{2}+y^{2}+z^{2}}}\label{EQUATION.10d1bacb-a0cf-4bdc-8b6d-c72d845b975b}$$

Then we find the satisfying result

$$\overrightarrow{r}^{\prime\prime}+2\widehat{z}\times\overrightarrow{r}^{\prime}+{\left(\widehat{z}\cdot\overrightarrow{r}\right)}\widehat{z}=\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\overrightarrow{\nabla}\mathrm{\Omega}\label{EQUATION.7aeaeb03-1226-46ab-815a-4b28e71a84a5}$$

The individual components of \eqref{EQUATION.7aeaeb03-1226-46ab-815a-4b28e71a84a5} are

\begin{aligned}x^{\prime\prime}-2y^{\prime} & =\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\frac{{\partial\mathrm{\Omega}}}{{\partial x}}\\y^{\prime\prime}+2x^{\prime} & =\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\frac{{\partial\mathrm{\Omega}}}{{\partial y}}\\z^{\prime\prime}+z\hspace{0.9em} & =\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\frac{{\partial\mathrm{\Omega}}}{{\partial z}}\end{aligned}\label{}

where

$$\begin{array}{rcl}\overrightarrow{\nabla}\mathrm{\Omega} & = & \left[\begin{matrix}x-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}\left(x+\mathrm{\mu}\right)-\dfrac{\mathrm{\mu}}{r_{2}^{3}}\left(x-1+\mathrm{\mu}\right)\\y\left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mathrm{\mu}}{r_{2}^{3}}\right)\\z\left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mathrm{\mu}}{r_{2}^{3}}\right)\end{matrix}\right]\\ \\& = & \left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mu}{r_{2}^{3}}\right)\overrightarrow{r}-\mathrm{\mu}\left(1-\mathrm{\mu}\right)\left(\dfrac{1}{r_{1}^{3}}-\dfrac{1}{r_{2}^{3}}\right)\widehat{x}\end{array}\label{}$$

## How I Do MathJax II. Example

To render equations in a WordPress blog, you have several options. The most aesthetically pleasing is MathJax. An earlier post tells you how to install MathJax for your WordPress site. This second post shows a few pointers by way of an example (you’ll probably want to view the page source, then search for “For example”). Here are a few more usage examples.

### How to Do Math in a Blog Post

If you’ve installed MathJax in your site, then in a blog post you can trigger the loading of MathJax by putting the shortcode at the top of your post. It will not show up in your readers’ browsers.

That’s it! You can write your post now.

What I usually do, if the document has a lot of equations, is to compose the post in the quasi-WYSIWYG LaTeX editor, LyX. You can, of course, use whatever writing tool you like. When you’re happy with how your article looks, then copy the text to the clipboard. (With LyX, open up the source pane (View→Source Pane) and select the text.) Paste to your WordPress post editor.

You now have to make one change to the pasted text: remove the line breaks inside AMS environments

\begin{...} ... \end{...}

For example,

$$\begin{array}[b]{ccl} D & = & -\left(R+h\right)\cos z\pm\sqrt{\left(R+h\right)^{2}\cos^{2}z+\left[\left(R+H\right)^{2}-\left(R+h\right)^{2}\right]}\\ \\ & = & \left(R+h\right)\left(\sqrt{\cos^{2}z+\dfrac{\left(R+H\right)^{2}-\left(R+h\right)^{2}}{\left(R+h\right)^{2}}}-\cos z\right) \end{array}\label{eq:D soln quadratic ugly}$$

becomes

$$\begin{array}[b]{ccl}D & = & -\left(R+h\right)\cos z\pm\sqrt{\left(R+h\right)^{2}\cos^{2}z+\left[\left(R+H\right)^{2}-\left(R+h\right)^{2}\right]}\\\\& = & \left(R+h\right)\left(\sqrt{\cos^{2}z+\dfrac{\left(R+H\right)^{2}-\left(R+h\right)^{2}}{\left(R+h\right)^{2}}} \cos z\right)\end{array}\label{eq:D soln quadratic ugly-how}$$

Here’s how to refer to the above equation. Write, for example,

eq. \eqref{eq:D soln quadratic ugly}

which renders as eq. \eqref{eq:D soln quadratic ugly-how}.

## How I Do MathJax I. Installation

I use equations. To enable equations in a WordPress blog, there are several options. The most comprehensive—and aesthetically pleasing—is to use MathJax. This post tells you how to install MathJax for your WordPress site. A second post has a few pointers. Here are a few usage examples.

#### 1. Edit default.js

I do not use the MathJax CDN since occasionally their site has problems. When that happens, your math stops working and your pages containing math become ugly. So I download MathJax to my WordPress install. Rather than futz with <script> tags in my site’s header, I edit the default configuration file to my liking. Thus:

• Unpack the archive file to your hard drive.
• Edit default.js in the config directory. My preferences:
• You’ll probably want to add to your extensions, something like:
extensions: ["tex2jax.js", "TeX/AMSsymbols.js", "TeX/AMSmath.js"]
• Scroll down and set messageStyle to your liking (I changed mine to messageStyle: "simple").
• Scroll down to menuSettings and change these to your liking (I set zoom: "Hover").
• In the tex2jax section that immediately follows:
• Under inlineMath uncomment the line with inline delimiters ['$','$']. This enables normal LaTeX inline delimiters. You’ll have to escape actual dollar signs with \\\\$.
• processEscapes: true
• preview: "[math]"
• Scroll down to the TeX section.
• Under equationNumbers, set autoNumber: "AMS".
• Fiddle with whatever else there catches your fancy.
• Fiddle with whatever else catches your fancy.