# reperiendi

## A first attempt at re-winding Escher’s “Ascending and Descending”

Posted in Uncategorized by Mike Stay on 2010 May 19

And he dreamed, and behold a ladder set up on the earth, and the top of it reached to heaven: and behold the angels of God ascending and descending on it.

Edit (May 20):

Even though it’s not a conformal transformation, this version looks better in a lot of ways.

Rather than cramming the whole picture into a single window frame, it presumes there’s a concentric set of these castles, each half as small as the previous, and built within its open internal patio. Doing it really well would involve extending the walls out to the edge of the outer wall that obscures them.

## I’m not this guy

Posted in Uncategorized by Mike Stay on 2010 May 18

## Faith

Posted in Uncategorized by Mike Stay on 2010 May 13

Blind faith is the best antisceptic.

## Tag clouds

Posted in Uncategorized by Mike Stay on 2010 May 7

Tag clouds are cumulonymous.

## Soliton

Posted in Poetry by Mike Stay on 2010 May 7

With a stroke, the pilot glides forward across the lake.
He does not know the names of the vortices cast off by his oar;
Neither is he known to the Sun.

Posted in General physics, Quantum by Mike Stay on 2010 May 3

This works best with small groups of about 5-10 students and at least thirty dice. Divide the dice evenly among the students.

1. Count the number of dice held by the students and write it on the board.
2. Have everyone roll each die once.
3. Collect all the dice that show a ‘one’, count them, write the sum on the board, then set them aside.
4. Go back to step 1.

A run with 30 dice will look something like this:

 dice number of ones 30 5 25 4 21 4 17 3 14 1 13 3 10 2 8 1 7 1 6 0 6 1 5 0 5 1 4 1 3 0 3 0 3 0 3 1 2 1 1 0 1 0 1 0 1 0 1 1

Point out how the number of dice rolling a one on each turn is about one sixth of the dice that hadn’t yet rolled a one on the previous turn. Also, that you lose about half of the remaining dice after about four turns.

Send someone out of the room; do either four or eight turns, then bring them back and ask them to guess how many turns the group took. The student should be able to see that if half the dice are left, there were only four turns, but if a quarter of the dice are left, there were eight turns.

If the students are advanced enough to use logarithms, try the above with some number other than four or eight and have the student use logarithms to calculate the number of turns:

turns = log(number remaining/total) / log(5/6),

or, equivalently, in terms of the half-life (which is really closer to 3.8 than 4):

turns = 3.8 * log(number remaining/total) / log(1/2).

When Zircon crystals form, they strongly reject lead atoms: new zircon crystals have no lead in them. They easily accept uranium atoms. Each die represents a uranium atom, and rolling a one represents decaying into a lead atom: because uranium atoms are radioactive, they can lose bits of their nucleus and turn into lead–but only randomly, like rolling a die. Instead of four turns, the half-life of U238 is 4.5 billion years.

Zircon forms in almost all rocks and is hard to break down. So to judge the age of a rock, you get the zircon out, throw it in a mass spectrometer, look at the proportion of uranium to (lead plus uranium) and calculate

years = 4.5 billion * log(mass of uranium/mass of (lead+uranium)) / log(1/2).

Problem: given a zircon crystal where there’s one lead atom for every ninety-nine uranium atoms, how long ago was it formed?

4.5 billion * log(99/100) / log(1/2) = 65 million years ago.

In reality, it’s slightly more complicated: there are two isotopes of uranium and several of lead. But this is a good thing, since we know the half-lives of both isotopes and can use them to cross-check each other; it’s as though each student had both six- and twenty-sided dice, and the student guessing the number of turns could use information from both groups to refine her guess.

## Escher and Mandelbrot

Posted in Math by Mike Stay on 2010 May 3

If you take a complex number z with argument θ and square it, you double θ. The Mandelbrot/Julia iteration

z ↦ z2 + c

does pretty much the same thing, but adds wiggles to the curve. Since the iterations stop when |z| > 2, the boundary at the zeroth iteration is a circle; after the first it’s a pear shape, and so on. We can map any point in the region between bands to a point in a rectangular tile that’s periodic once along the outside edge and twice along the inside edge. Here’s a site with a few different examples.

The transformation Escher used in “Print Gallery” takes concentric circles at r=1/rn to a logarithmic spiral. The concentric boundaries between iterations for the Julia set at c = 0 are circles. There ought to be a transformation for Mandelbrot / Julia sets similar to the Droste effect but spiraling inward so that the frequency is smoothly increasing just as the distance can be made to smoothly increase.

## The animated gif of Dorian Grey

Posted in Uncategorized by Mike Stay on 2010 May 3

My friend Jas wrote a Perl module, Perl::Visualize, that makes Perl/Gif polyglots. He later adapted his technique to Javascript/Gif polyglots. Some guy generalized a quine to print out the source code of the program together with a comment containing the next frame in Conway’s Game of Life. So we have programs that can age themselves, as well as programs that double as pictures. The “aging” process can repeat. So someone make a JS/Gif polyglot quine that renders consecutive frames of Dorian Grey!