It’s really quite amusing to repeatedly apply a series of effects to an image, say using Photoshop filters. I didn’t think about it until yesterday, but this is a great way to model various kinds of things, particularly certain partial differential equations.

To take a really simple example, we all know that heat diffuses, and that in the absence of any energy being pumped into a system, temperatures will tend to even out over time. Similarly, if we apply a blur to an image over and over again, soon we will have a smeared out gray. 

Heat is modeled in a continuous realm: time is continuous, space is continuous. But while blurring, time occurs in discrete steps, and space is modeled in discrete pixels. Nonetheless, the two phenomena are closely related.

The famous heat equation is very simple:

du/dt   ∝ Δu  

All this means is that the amount of a quantity ‘u’ in a given location changes over time, and this change is proportional to how much net  variation there is nearby– the more variation the more u will change. (For those with more than Cal III under your belt, no need to explain; if you’ve had Cal I, if we measure temperature along a rod, then this Δu works out to be just d²u/dt² and sure– the more concave up the function is, say, the faster the heat will increase at that spot, in an attempt to smooth out the temperature; if you haven’t had any calculus, or even if you have, it’s interesting to consider a discrete version of this:

Suppose we have a bunch of graph paper, with a temperature written in every cell. Then the new temperature at each new time can be modeled by taking a weighted average of the neighbors. For example, if we’re at X :

a  b  c

d  X  f       

g  h  i

we might take an average using none of a,c,g and i, 1/8 of b,d,f,h and 1/2 of our original value X. (Generally we count closer spots more)

i.e. new value = X/2 + (b+d+f+h)/8

The change from the old value to the new value is —X/2 + (b+d+f+h)/8, and sure enough this is a discrete form of the laplacian.

We used photoshop to make the image at left , at the top of this post: we’ve applied a gaussian blur (rad = 1 px) and then three sharpens, and then repeated this over and over again.

And look!! Unmistakably, the spots and stripes that are the hallmarks of reaction diffusion types of equations have popped up!!

Reaction diffusion patterns pop up in many places– not least of which in the patterns of spots or stripes on many living things. The idea in reaction diffusion is that one or more quantities (like the amount of black in a pixel, or the amounts of various hormones or chemicals) simultaneously diffuse (blur) and react (changing the concentrations). These two actions are in a kind of tug-of-war, and when they are well-matched all kinds of interesting things can occur.

Here’s my question (for experts only): that doesn’t make sense– aren’t both a gaussian blur and sharpening modeled by adding / subtracting a laplacian? And where is the reactive term?

Incidentally, here’s another example that works beautifully:

If we repeatedly apply a gaussian blur and heighten the contrast, we obtain a sequence of images as below

This is exactly what we should expect: this is a discrete form of a well-known reaction diffusion equation in which the reactive force drives the interface between black and white towards having less and less total curvature. 

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Incidentally, the discoverer of this class of  equations was none other than the great Alan Turing, father of the theory of computation (and, as it happens, leader of the effort to crack the Nazi enigma code).  I  must take the opportunity to remind us all again that this great man was hounded to his suicide for his homosexuality by the very government he’d worked to save— a clear a martyr to the cause of universal human rights as there can be.

I was wondering what is the theoretical ‘area’ of contact between two spheres in contact with each other. I was unfortunately not able to locate much (if any) information on this. After some thought into this I’ve realised that the spheres would meet at a single ‘point’ however what would the area of this ‘point’ be? The only source related to this claimed the area of contact, the point, has no area. How can a point have no area? If the spheres touch, musn’t there be an area shared between them? Even if only one atom?

Hi, the issue here is that there is a vast difference between physical, real things and the mathematical ideas that model them.

Real, mathematical spheres don’t exist, plain and simple! Never could, even as a region of space— space itself has a granularity (apparently) at a scale of about 10^-33 meters. There simply cannot exist a perfectly spherical region in physical space, much less a perfectly spherical body.

But as an abstraction, the idea of a sphere is very useful: lots of things, quite evidently, are spherical for all practical purposes.

For that matter, “points” don’t exist either, and are also a mathematical abstraction. (So, too, is “area”. Real things are rough, bumpy and not at all like continuous surfaces, on a fine enough scale) But again, these _ideas_ are very good at getting at something important about lots and lots of physical things, and so have proved useful.

Tangent spheres do indeed meet in a single point, which has no area.

Spherical things meet in some other, messier way.

Hope this helps!

Mark A. recently wrote us:

The contestant started with 25 cases each with a value of dollars inside. They select one case with the hopes that it has the “Grand Prize” (GP) amount inside. Cases are opened and other amounts eliminated. In the end only two cases remain. The contestant has the one they selected and the other is still in the “Game Field” across form them. One case has the 1,000,000 dollars GP and the other has 10 dollars. As I understand it there is a higher probability that the GP is in the case not selected by the contestant in the beginning so that contestant should always swap cases. I do not see how this can be so. I understand that in the beginning the case selected only had a 4% chance of having the grand prize (GP) but so did the other remaining case. As cases were eliminated the odds of these cases having the grand prize raised at the same rate to the point that now they should each have a 50% probability of having the GP. In addition in the beginning each case also had a 4% chance of having the 1 dollar and that chance also rose to 50% for each case. I do not see any preferred state where the decision of the contestant has an effect on the distribution of the odds.

I see this interpretation of Bayesian analysis often and I can not see the validity. Am I wrong?

This is a really fascinating question, and the answer depends on the critical phrase:

…Cases are opened and other amounts eliminated.…

Strangely, it depends on WHO is doing the eliminating and with what knowledge!

In other words, are they eliminated in such a way that the GP must be in play at the end, or in such a way the game might have been aborted prematurely?

A) If the contestant chooses, or the cases are chosen randomly (i.e. the GP was at risk at every stage), then the probability is the same for each case, at each stage, right to the end. It doesn’t matter either way if the contestant switches. This is the way Deal or No Deal is played.

B) If the game-show host, or some knowledgeable party removes cases from play (knowing they do not contain the GP), then it is better to switch. Incidentally, this version is known as “The Monty Hall Problem”, after the host of the 70’s game show Let’s Make a Deal (In which a contestant would be offered three doors, one of which conceals a fantastic prize; the contestant chooses one door, and then Monty Hall would eliminate one of the remaining doors that doesn’t have a prize; the contestant is then given a chance to switch to the last, unopened door— an opportunity which should always be taken!)

This seems paradoxical, doesn’t it? The knowledge and intention of the person removing cases from play seems to change the probabilities.

But this really does make sense.

—-

In (A) the probabilities remain equal, in effect, because no action has been taken that changes the relative likelihood of any outcome. Suppose we have, at a given stage, N equally likely possibilities, and one is removed at random, if the game does continue (which it might not) then there now (N-1) possibilities— all of which are still equally likely, etc.

In (B) the actions change the relative probabilities. This is a little harder to explain, but in a nutshell, the host sweetens the deal: your original choice is just as likely to hold the prize, but the other choices have become more likely to be winners, since a losing choice has been removed. Let’s count out the possibilities:

Suppose we have three briefcases are a, b, c, and the prize is in case a. We will list them in the order of

“case chosen by the contestant, case eliminated, case remaining”

The game would have ended if case a had been eliminated, so this leaves only

a b c (contestant should keep)
a c b (contestant should keep)
b c a (contestant should switch)
c b a (contestant should switch)

In (A) each of these is equally likely, since each of the choices was made completely at random. Any of the six sequences
a b c
a c b
b a (stop)
b c a
c a (stop)
c b a
was equally likely (Since there is 1/3rd chance the contestant will pick a,b or c; then there are two equally likely possible ways for one of the remaining case to be eliminated; the final case, if there is one, is determined)

Now 1/3rd of the time the game ends prematurely, but if the game finishes, there is 1/2 probability that the contestant should switch– it’s 50-50 either way.

In (B) though, the choices are not equally likely.
There is still a 1/3rd chance that the contestant chooses a, 1/3rd b, 1/3rd c.
If the contestant chooses a, it is equally likely that the host opens case b or c;
On the other hand, if the contestant chooses b, the host will certainly open c; if the contestant chooses c, the host will certainly open b.

so this gives

a b c 1/6th of the time
a c b 1/6th of the time
b c a 1/3rd
c b a 1/3rd

2/3rds of the time, the contestant is better off switching.

m choose i = m!/(i! (m-i)!)
m choose j = m!/(j! (m-j)!)

for all counting numbers i,j,m with 1 < i,j < m Another way to state this same thing is: any pair of entries, on any row of Pascal's triangle (except for the 1's on the edges) will have a common factor. With facts of this sort, often there is a clever way to cast things in terms of counting something a couple of different ways which makes things clear.

I was listening to another podcast and they misread the copy and ended up
saying “What is the most numerous number?”. Well, what IS the most numerous number?

This is really a fascinating question! Have you ever wondered, for example, why there are 7 of so many things:

• 7 wonders of the ancient world
• 7 mortal sins
• 7 stars in the big dipper
• 7 days of the week
• 7 dwarves
• 7 brides for 7 brothers
• 7 items on this list

Really, it’s not that big of a mystery. The fact is, small numbers are very useful, and get called upon a lot. But there aren’t that many of them to go around.

Hence, the First Strong Law of Small Numbers: There aren’t enough small numbers to meet the many demands placed upon them!

The most numerous numbers, in a sense then, are the small ones. Google searches seem to confirm this:

• 1, 2, 3, 4, … (several billion hits each)
• 78, 122, 157, … (several hundreds of millions of hits each)
• 12122…(millions of hits)
• 1278232… (hundreds of hits)

Lotsa fun can be had in this way… With a little fishing, you can find some ridiculously large numbers with more hits than they deserve, but the principle is clear.

This same principle, incidentally, explains why, for example, the Golden Ratio appears in so many settings. There’s nothing really that mystical about it. The Golden Ratio is a root of the polynomial x²-x-1=0. Roots of polynomials come up all over the place, in countless applications. And just as small numbers are in great demand, roots of simple polynomials will appear over and over again.

The Golden Ratio is just about the simplest non-integer root possible, and so, of course, shows up endlessly.

Challenge Question I’m kind of curious now: What is the smallest counting number that is NOT on the web?

210210876 was not on the web until just now, according to Google. Internet history has just been made!! But I’m sure you can find something smaller…

In this corner we have
Sequence 1 n^^n
1, 2^2, 3^3^3, 4^4^4^4, and so on.

And over here we have Sequence 2, defined recursively by

In short, we can define the second sequence as s(1) = 1; s(n) = n, followed by s(n-1) factorial signs.

Which sequence grows faster than the other??

We have many conflicting answers, and no decisive resolution; here was one idea .

Have you ever discussed factorials on your podcast? I don’t recall,
but a friend and I are puzzled and so of course we turn to you: Why
is “zero factorial” 1? Was it simply defined that way to frustrate
all of us nonmath folks, or is there a valid explanation?

Well, David, I think it’s fair to say it was made up, but for a very very good reason. Many formulas involving factorials are simply more uniform if 0! = 1.

Here’s one way to think of it. 4!/4 clearly gives 3!.
Similarly 3!/3 gives 2!
2!/2 gives 1!
So of course, 1!/1 should be 0!, and so it is!
(But then (-1)! is 0!/0, which doesn’t work well at all!)

Combinatorial questions give another way to think of it. The number of ways to order n objects in a line is n!. How many ways are there to order 0 things? It’s not impossible (which would be 0 ways)—there is 1 way to do it! — don’t do a thing!

Similarly, the number of ways to marking k objects in a pool of n objects, say with a red crayon, is n! / (k! (n-k)!) This is the same as not marking the other (n-k) objects red, but, say, blue instead. Clearly there is one way to mark all the objects blue (i.e. mark none of them red). The formula only works if 0! is defined to be 1.

Another answer, closely related, is that n!, as a function of n, can be smoothed out and made continuous, by something called the Gamma Function. Gamma(n+1) = n!, and as it happens, Gamma(1) = 1 so 0! should be 1. (I bet you would never have guessed that (1/2)! is sqrt(pi)/2; well it isn’t, really, but Gamma(3/2) = sqrt(pi)/2

But the bottom line is, it’s really a matter of making a wise definition.