The relevance of Pascal’s Triangle to a “best of seven” tournament, such as the World Series, is obvious. But there is that wrinkle of not actually playing all seven games. In the past I had contented myself with various adjustments, but I never actually drew up “Pascal’s Diamond”, which is based very simply on the same generating rule as the Triangle, but terminated appropriately, as shown here:

Note that it’s equivalent to Pascal’s Triangle up to game 4 ( as indicated by the black numbers ) but terminates along the diagonals indicating 4 wins by either team. We immediately see that the nominal probability ( assuming “chance” outcomes ) of a 4,5,6, or 7 game series is 1/8, 1/4, 5/16, or 5/16, respectively.

Note that the equal probability of a 6 or 7 game series is a reflection of the “chance” outcome of game 6, which terminates the series at 4-2, or forces a game 7 with equal probability.

I decided to compare these probabilities with the last 91 world series results, in terms of games played, or equivalently, “games won by the series loser”. I found these to be 18, 18, 20, and 35. This compares to a nominal expectation of 11.37, 22.74, 28.43, and 28.43, for 91 “chance” games. So, this looks a little out of whack! But is it really? …

Oh No! Not STATISTICS!

Well, we don’t have to do a canned analysis, even if we find ourselves being driven that way. We can make up any sort of test and apply it experimentally to sets of 91 “chance” series, and determine the “probability of rejecting the null hypothesis”, in this case the hypothesis that the real outcomes obey the same statistics as “chance” would have it.

Let’s just try looking at a generated sequence of 91 chance-ruled series. Here’s 10 of them, with the average over all 10 at the bottom. Each row represents the number of 4,5,6, and 7 game series in a trial of 91.

$ series 10
9 20 31 31
9 18 38 26
12 19 32 28
14 14 37 26
12 20 31 28
12 19 33 27
12 24 32 23
11 23 27 30
11 25 27 28
8 26 26 31

11.000 20.800 31.400 27.800

Well, notice the 14, 14, 37, 26 … qualitatively very similar to the actual historical “trial” of 91 world series, so impressionistically, we don’t have grounds to believe the games are ruled by anything but chance.

I’ll just add that I did go “back to the books”, or Wikipedia as we do these days, and tutored myself on some of the sigma-based statistical tests, and I’m ready to report that it’s all very interesting!

In Chapter IX of ALICE IN WONDERLAND Alice has a conversation with “the Duchess”, wherein the Duchess relates to her several “morals,” including the following:

“Be what you would seem to be”— or if you’d like it put more simply —“Never imagine yourself not to be otherwise than what it might appear to others that what you were or might have been was not otherwise than what you had been would have appeared to them to be otherwise.”’

It has occurred to me, and I feel firm in this, that Lewis Carroll meant this as a mock paraphrase of Proposition 6 of Book II of Euclid’s Elements:

If a straight line is bisected and a straight line is added to it in a straight line, then the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half equals the square on the straight line made up of the half and the added straight line.

I forget the exact occasion, but it was as I was perusing Euclid that I was struck by the similarity of construction. The phrase, “straight line” , corresponds roughly to “otherwise” and “others” in the Duchess’ moral, and both expressions suggest a lilting meter. ( … and I hope I don’t have to remind anyone of Charles Dodgson’s mathematical vocation ! )

Furthermore, the syntax is similarly baffling, Euclid can actually be parsed, but not so the Duchess. Her moral contains three long phrases, each coherent on its own but with overlapping connections that make the whole thing incoherent:

Never imagine yourself not to be otherwise than what it might appear
to others …

it might appear to others that what you were or might have been was not otherwise than what you had been …

what you had been would have appeared to them to be otherwise.

However, note that Prop. 6 does contain a “garden path” , as described by Stephen Pinker in THE LANGUAGE INSTINCT.

This is the phrase, “the rectangle contained by the whole with the added straight line …” which one is inclined to take as a complete description of a rectangle. However, the following phrase, “… and the added straight line together with the square on the half … “ is rendered nonsensical if one goes down this path, because a line cannot be taken together with a square. Of course, (ahem) the phrase, “with the added straight line” refers to the base of the rectangle while, “and the added straight line” specifies the height.

To me, it all adds up, especially considering the preamble. “Be what you would seem to be” would stand for the algebraic expression:

(a+b)² = a² + 2 a b + b² … OR … a ( a + 2 b ) + b²
… as per the theorem.  ( Note the “bee” and “two bee” in the moral. )

Also, “If you’d like to put it more simply”, alludes to the historically primitive character of geometry vis a vis algebra, which is somewhat more abstract. So Euclid’s book II may be construed as simpler than modern algebraic notation, being put in more primitive, or “simpler” terms.

Well, Heath was 40 years after ALICE, but I think that the english version of basic Euclid was probably already a matter of tradition by this point, although I really couldn’t say, of course.

NOTE ADDED IN PUBLICATION

All the above I’ve been carrying around for some years, but in anticipation of writing this up, I was spurred to check the original greek of Proposition 6:

… so feast your eyes! heh heh. Well, I spent some time on it but I don’t want to belabor the issue. You can note “tetragono” ( square ) and “gramme” ( line ) and “orthogonion” (!) orthogonal.

The thing I want to say about it is that the “garden path” does not exist in the greek, and the syntax seems to be a little more technical. “apo” means the base of a square, so it announces the specification of such, which is terminated by “tetragono” so that’s it, with no ambiguity.

The “garden path” clause specifies the base and height of a rectangle delineated by TO [ base spec. ] KAI [ height spec. ]

The base spec. is “upo tes holos syn te proskeimene” ( base of the whole with the extension ) and the height spec is “proskeimenes periexomenon orthogonion” ( encompassing orthogonal extension ? ) then comes “meta” = “together with” … ( the square on the half ) so there seems to be a more elaborate system of “terms of art” in play.

… of course I allude to the punchline, “Consider a spherical horse”, although there is such a thing as a spherical square. Namely a quadrilateral drawn on the surface of a sphere with equal sides and equal angles. Many years ago I was drawn into contemplation of this by a statement in Peterson’s Field Guide to the Stars and Planets, by Donald Menzel, that “The sky contains approximately 40,000 square degrees …” as excerpted here:
A more accurate value is given by 4Pi/(Pi/180)²  = 41252.96125, but this is still approximate, as it does not allow for the curvature of the “square degree” itself. So I endeavored to derive an exact value for the area of a spherical square degree, and did so, as recorded in this note:

I was quite pleased with this result at the time, as it was simple and even had a certain elegance, and in fact I had difficulty reproducing the calculation before I rediscovered my note. I did remember the result verbally: “four arcsine of sine squared alpha over two”, where alpha is the measure of the square across the center.

Applying this formula gives the value of 41254.00842 for “square degrees in the sky”, differing by about 1 in the units place from the linear approximation.

I recall that I knew at the time that the formula was “four arcsine of tan squared beta over two”, where beta is a side of the square. This gives 41250.86683, undershooting the linear approx. by about twice the amount that the alpha formula overshoots. I’m sure there’s a story there, but let’s move on.

I was recently drawn into contemplation of the spherical square by a NYT Quote-acrostic based on a quote by Calvin Trillin, which alluded to the infamous Indiana Pi Bill ( misattributing it to Texas ). The cited Wiki article offered in explanation of the bill’s content, a crank mathematical screed, that “it is clear that the assertion is simply that the area of a circle is the same as that of a square with the same perimeter”

I started wondering if this might not be true for some spherical square and circle. I finally realized that it is true of the “degenerate” spherical square whose sides are quadrants of an equator, but too late! … I was back on the case.

In my calculation, I had retreated to the comfort zone of spherical coordinates, and resorted to integral techniques which, while well within the bounds of Introductory Calculus, were less than straightforward.

I knew that the problem could be solved by methods of Spherical Trigonometry, but I didn’t see that these were any simpler.

Well, this time around I came up with a simple construction that makes it a “one step problem” , applying the rule of “spherical excess” for the area of a spherical triangle:

Well OK, a “two step problem”, we still have to solve for theta. This is an easy application of Napier’s rules for right spherical triangles applied to either the light blue or lavender auxillary diagram in the figure, and giving

or …

Additionally, we may solve for the distances x and y, which are the half-lengths of the sides of the spherical rectangle. Applying Napiers rules, as above, we get:

Note that when psi = chi and equivalently x = y, we have the identity

… as alluded to above.

I mentioned the image gallery section of the Wiki Mandelbrot Set article as the starting point for my renewed interest in this topic. It is based on a zoom sequence terminating on a “Julia Island”, which is a component of the Mandelbrot set resembling a “Julia set”, which is based on the actual sequence of an iteration rather than the in/out criterion of the Mandelbrot set. So there’s a little math magic in this resemblance.

Here is my recreation of the zoom, based on the linked tool I mentioned, which by no means outstrips the images in the article ( which can be expanded, ) but it does go a little deeper, and it keeps the Julia Island location at center. This was easy to do by working backwards from the deepest image and keeping the same coordinates by simply changing the scale in the URL field.

Here’s a “hot link” of the lowest level of the Wiki zoom :( To see its full glory visit the site. ) It roughly corresponds to the antepenultimate frame of the animated gif. The commentary in the main article notes that there is a “satellite” ( i.e. “mini-Mandelbrot set” ) at the center of the island structure that is “too small to see at this magnification”. Just visible there is the large square structure in the center of the last frame of the gif, where you can almost see the satellite at the center of it, as a tiny dark speck.

This conclusion is supported, if not confirmed, by perusal of a nearby and entirely analogous structure, which is a little bit larger. Here’s a slower zoom, taking 1/2 per frame instead of 1/10, showing a descent through the analogous square structure to reveal the Mandelbrot satellite.

Note the emergence of a new “sea” of similar spirals and structures as part of the INFINITE descent. It’s fascinating and incredible and, like the song says, “a little bit frightening.”

I recently saw one of several “Deep Zoom” videos of the Mandelbrot set, which I’ll leave alone for now, but it got me interested in the subject, which I haven’t thought about since the 1980’s. Times change, and the facilities available for investigating the subject are now, of course, much more powerful. I looked at the Wikipedia article on the Mandelbrot set, which refreshed and greatly expanded my casual knowledge of the subject. The exciting thing though is the link under the section headed “Image gallery of a zoom sequence” to an “interactive viewer” ( go to Wikipedia to get the link. ) It indicates it will open for “this location” ( i.e. the one illustrated ) but I found it goes to the initial full view of the set.

Anyway, it’s a fabulous tool. You can click on a spot to zoom by a factor of two, and although it does take time to calculate the image, you can click again for repeated zooms without waiting for the full resolution. I found it very handy to use. It goes down to 10^-14 in scale, which is well short of the claims made by these deep zoom videos, but it provides ample scope for casual exploration.

Note the parameters are passed in the URL address field, and these can be edited and entered with the naturally hoped for result. Of course you can save them and reenter them to bring back the image, so as I say, very handy.

I also used the extended precision calculator, bc, on Cygwin to explore some of the mathematical aspects, which I’ll get to. First, here’s an animated gif of a short zoom on a large portion of the Mandelbrot set, progressing down the scale of the circular “bulbs” along the real axis. This is a more extensive version of the “Mandelbrot zoom” gif in the Wiki article which uses a loop of two circles without “hair”.

First note that the Mandelbrot set itself is the black area, determined by points c, where the iteration of the defining formula always remains bounded. The “hair” is exterior to the set and is colored according to how rapidly the formula diverges under iteration.

These circles are in the set, and are called “period bulbs” in the article, because of the defining nature of the points within them. That is, they are points c, in the imaginary plane, such that iteration of x = x^2+c ( where ‘=’ means assignment, as in a programming language ) starting with x == 0 ( ‘==’ asserts equality ) converges to a cycle of fixed values, these values being 2 in number in the “2” bulb ( the first circle ) then 4 in number, then 8, 16, 32, and 64 in number in the large circle of the sixth and last frame.

This brings us to the math, and bc. Let’s go for the gusto with the “64 bulb”. Using the “click and zoom” facility in the interactive viewer, we go to the sixth bulb, i.e. the 64 bulb, and simply copy and paste the x-axis ( i.e. real part ) value into bc, and initialize x to 0:

c= -1.400968738408754
x=0

Then we iterate for a goodly number of times to assure ourselves of convergence to a stable cycle, in the belief this must happen:
( Even though bc runs interpretively, the power of a modern PC is such that this number of iterations causes a barely perceptible hitch in the response )

n=100000
for( i=0 ; i<n ; i++ ) x=x^2+c;x
.010561399880314

Now we check for a 64 cycle of values:

n=64
for( i=0 ; i<n ; i++ ) x=x^2+c;x
.010561399880314

OK! But maybe this is a repetition of a shorter cycle, so check 32:

n=32
for( i=0 ; i<n ; i++ ) x=x^2+c;x
-.000137388541270
for( i=0 ; i<n ; i++ ) x=x^2+c;x
.010561399880314

OK again!  We can now be satisfied that 64 is the true cycle value. Note that we didn’t use a lot of precision, but the “attractive cycle” is strong medicine, and we don’t need it. We are just checking the fundamentals though, and of course there are more delicate arrangements to be found.

In particular, what about the “vanishing point” of this self-similarity? It is cited in the article to crude precision as -1.401155, which can be discerned numerically, but what are its defining properties? This point remains obscure to me.

Misiurewicz points

These are defined as values of c such that k iterations of x=x^2+c give a value that repeats after n ( more ) iterations, and are designated as M_k,n, and these form vanishing points for self-similar transformations although they do not account for the seemingly fundamental one just described. Noteworthy is M_4,1  described in the Wiki article.

Here is a zoom which goes by steps of 100X magnification ( after an initial step of 10X )
The Wikipedia article on Misiurewicz points mentions … “(Even the `principal Misiurewicz point in the 1/3-limb’, at the end of the parameter rays at angles 9/56, 11/56, and 15/56, turns out to be asymptotically a spiral, with infinitely many turns, even though this is hard to see without maginification.)” ( sic ! just noticed that. ) … This is our guy, and you can see the “slow roll”, which advances by about 50 degrees in 6 steps of 100X magnification. ( I’m still working on the meaning of those angle values. )

In this case we can do a little math and come up with a way to define the M_4,1 point, at the center of the zoom, to arbitrary precision. Approaching it graphically we see the center at approximately ( -.10109636 , .95628651 ) in the complex plane.

In bc we define functions:

define f(a,b){ return( a^2-b^2 + c); }
define g(a,b){ return( 2*a*b +d ) ; }

… which are the real and imaginary parts of the formula, x^2 + (c,d) , applied to a complex value x with real and imaginary parts (a,b) . So we can proceed to apply the formula iteratively and see if we come up with a constant value after 4 iterations, as per the M_4,1 designation:

c= -.10109636
d= .95628651
x=y=0
z=f(x,y);y=g(x,y);x=z;x;y # iteration 1
-.10109636
.95628651
z=f(x,y);y=g(x,y);x=z;x;y # iteration 2
-1.0053597752027305
.7629323394437928
z=f(x,y);y=g(x,y);x=z;x;y # iteration 3
.32758616302650612570470420629841
-.57775646055620961839245967248080
z=f(x,y);y=g(x,y);x=z;x;y # iteration 4
-.32758619350801035155822390304043175425331545129536
.57775646584523271983382407623684575759993010598581
z=f(x,y);y=g(x,y);x=z;x;y # iteration 5
-.32758617964890595894029463234712218921204878613363
.57775642715823884156305664636678518208809205705010

Note that the 4th iteration approximately achieves the repeated value, but the 3rd iteration is the negative of it. So to home in on the exact value we want to adjust c so that f(f(f(f(0)))) = – f(f(f(0))). Without spelling it all out, we can do this by numerically evaluating the sum of the 3rd and 4th iterate ( which we want to be 0 ) to see how it varies with small variation in the real and imaginary parts of c . This is essentialy “Newton’s method” of finding roots. Inverting a simple 2X2 matrix tells how to vary ( c, d ) to bring this sum to 0. By iterating this estimation we can find  the correct complex value of M_4,1 to arbitrary precision.

Here’s as far as I’ve taken it so far:
( c, d ) = (
#real part
-.1010963638456221610257854457386225654638054428262534
838769311776607808407404705842748212198105167791
,
# imaginary part
.9562865108091415007710960577299774358098333365105291
700343143215005246590657167325269784107873398073
)

It’s very easy to do and could be taken to many thousands of decimal places in less than an hour, I’m sure.

I’ve got a lot more, but I want to put something out now and I do plan to continue my discussion … but let me just outline my thoughts.

The mathematical structure of the Mandelbrot set has been a matter of ongoing investigation, and is a difficult subject. One fundamental point is the “connectedness” of the set, and it has been proven that it is “connected” according to the literature. Well, I’m sure it has, but I still have questions! In particular it seems to me that these Misiurewicz points must be isolated elements of the set. Perhaps in some sense they are, since topology has more definitions than you can shake a stick at.

Anyway, I hope to be back with more pictures.