I Don’t Think That Model Means What You Think It Means…

This is something I debated a long time whether or not to actually write anything about, mainly because the subject matter is not something that interests me, but the situation and story behind it is certainly something worth thinking about. Neverthless, this will likely be far too many words on a matter that deserved much less.

Thinking about population models, a relatively basic model would be to assume that the rate of growth of a population is proportionate to the size of the population. That is, at a given time t, for some characteristic constant k,

That constant encapsulates a lot of information about the population, such as gestation rates, reproduction rates, etc.

Skipping the mechanics of actually solving it, you get the following,

Where P₀ is the population size at time t = 0.

Now, this solution allows us to answer an interesting question, given a population – how old is that population? How long must it have been in existence to reach the population size it is at now? The model is predictive forwards and backwards, and we can project when the population was at a given size.

We assume that the population had to start at some small value, say P = 1 for an asexually reproducing organism, P = 2 for a sexually reproducing organism, and P = 3 for a hedonistic, liberal organism. We’ll assume sexual reproduction. Then we simply solve for the time at which the population would’ve been that size.

We can then say that the age of the population is whatever time it is currently, minus the time given above. Which is all very interesting.

“Well, Fox,” I hear you say, “that’s all well and good, but where’s the proverbial beef?”

Oncoming Storm…

So much to talk about, so little time to do it. On the cusp of the last two weeks of classes here, so that’s fun. Exams soon, which will be really rather special. Many interesting math things to be talking about. I’ve got loads of things I’d like to show you how to count, using linear algebra, but first I’ve got to explain just what linear algebra is : ) And I’ve got some graph theory I’m working on, and an evolutionary programming project, and a bit on groups as I’ve been trying to get to for just ever now v.v

But if it must wait, then it certainly will. Things are going to get rough, and the only possible response is to get rough right back. They’ll bring it on, and I will bring it right back at them. The only way to meet the storm is to become a storm yourself.

And it’s going to be … fantastic.

Neat Trig Identity

Consider the following,

Action Stations!

Well, I’d like to talk about group theory and prime numbers, but it’s late and I’m tired, so I think I’m just going to talk about schedules and academic plans and things. Primes and groups may be a project better suited for the weekend.

So, it’s that time of the semester again. Classes are posted for next year, which normally means it’s time to hide from Kunal Ghosh, but this time it’s going to be different!

Because, much like the cylons, I have a plan.

Integer Partitioning: 3-Partitions

This will be, I think, a relatively gentle treatment of what could be a rather complicated matter.

The problem came up in problem seminar this week, the matter of integer partitioning. Specifically, given an integer n, how many different ways can you write it as the sum of smaller integers? More specifically, and a notably easier problem, how many different ways can you write it as the sum of three smaller integers?

For example, 6 can be written as 1 + 4 + 1, 3 + 2 + 1, and 2 + 2 + 2. Notice that we’re counting 1 + 4 + 1 as the same as 4 + 1 + 1. 5 can be written as 1 + 1 + 3, 1 + 2 + 2. So the number of 3-partitions for 6 is 3, and the number of 3-partitions for 5 is 2. Clearly 0, 1, and 2 all have no 3-partitions, and 3 has one, 1 + 1 + 1. But for a general n, how many are there?

The basic approach is going to be to come up with a method of constructing all 3-partitions of a given number, and then the formula for the total number falls out almost automatically. What follows then is a reduction from the initially simple to understand but difficult to compute formula to a relatively difficult to understand but surprisingly simple formula. It’s interesting.

Grab Bag: Busy Fox!

I’m being a busy, busy fox, and don’t have time for much high quality math right now. But I would like to point out two things, that Charles Daney has a good post up at Science and Reason, elaborating more on just what a transcendental L-function really is, a topic I’ve attempted to discuss before.

The other point of interest is that this is a really, really big rabbit.

Progress On Riemann?

The Riemann Hypothesis is one of the greatest unsolved math problems in the world today – probably. If not, it’s easily one of the most famous.

It centers on the Riemann Zeta Function, given by

The function being defined over the complex plane. The Hypothesis is, in its most basic form, that if , then s = 1/2 + t*i, for some real number t. That any zero of the zeta function has a real part of one half.

Why this is interesting though is the connection with prime numbers. There’s a function called the prime counting function, pi(x), and it basically gives the number of prime numbers between 0 and n. As the distribution of primes is so bizarre, being able to calculate pi(x) correctly, without just counting all the prime numbers from 1 to x, would be a very neat trick. We have some very nice approximations for pi(x), involving integrals of relatively simple functions, but the question is the, how good are these approximations?

So enters the Riemann Hypothesis. Basically, it’s been shown that the Riemann Hypothesis, as stated above, is equivalent to saying, for our best approximation of pi(x), pi(x) differs from this approximation by no more than some constant times , for x sufficiently large.

Which is a very interesting result.

But a proof was proposed a few months ago, and it seems really quite promising, and at the very least opens up some new avenues for further inquiry. As usual for such things, I don’t have the background to explain fully, but I think I can break down a couple of the key points to an understandable level. Maybe Mark can pick this up?

Prime Polynomials

Primes are of interest. Indeed, I would go so far as to say they are the stereotypical Object of Interest people think of when they think about mathematics.

When writing posts like these, I’m never really sure how basic to go. I’ve been tempted at times to define what functions are, for example. One of my goals in writing about math is to make it as accessible as possible, and I’m pretty sure I fail a lot at that, but then again I don’t want to go so basic I can’t talk about anything of interest.

The point is, it will satisfy me enough to say, primes are integer numbers that are only divisible by themselves and 1. 2, 3, 5, 7, 101 are classic examples, while 10 is very much not prime.

Primes are of interest for many reasons, but they tickle the fancy of many due to their rather inexplicable nature. For example, the distribution of primes, where they fall on the number line, while being highly structured, seems largely without form or pattern. We can’t really predict what numbers are prime, where they are, or how many there are in a certain range.

Something that would be nice would be a function that generates just prime numbers. Now, given a set of prime numbers, like 2, 3, 5, 7, you can construct, for example, a polynomial f(x) such that f(1) = 2, f(2) = 3, f(3) = 5, f(4) = 7, and so on. An excellent example of such a thing is f(n) = n² + n + 41, which produces primes for n = 0 to n = 40. Do you see why f(41) is not prime, though?

So the question is, in general, can you construct a polynomial that only gives prime numbers on integer inputs?

Too Many Words On Divisibility

In this one post I’m working on, of primes and polynomials, there’s this one part of the proof which requires a small lemma. And this has caused no end of vexation for me.

The problem is that the lemma is not directly related to what I’m trying to prove, and is not trivial enough to simply state. So it requires a proof, but were I to prove it, it would be this block of text in the body of my main proof completely unrelated to the point and topic I wanted to focus on. When I write these posts, I really try to keep everything focused and to the point to minimize the chance of losing anyone along the way. Moreover, while it’s a lemma about divisibility and remainders, a subject most easily communicated in the form of modular arithmetic, I don’t honestly know how many people who read this have exposure to that subject, so how much would I have to say to build up to the point of proving this lemma in that form? All this goes to say, how can I make the whole proof most accessible?

It’s a style and communication thing.

So ultimately, I’ve decided to let this one lemma, however simple, to have its own post. But, as a result, I will end up shedding far too many words on far too small a result.

The desired result is this, that for a prime p and two integers m and n, (1 + m*p)ⁿ gives a remainder of 1 when divided by p.

If this seems trivial to you, move on and be merry. If not, check below the fold. It’s a good divisibility proof, with some basic induction to boot.

Mathematicians Are Best!

“There are three men on a train. One of them is an economist and one of them is a logician and one of them is a mathematician. And they have just crossed the border into Scotland (I don’t know why they are going to Scotland) and they see a brown cow standing in a field from the window of the train (and the cow is standing parallel to the train). And the economist says, ‘Look, the cows in Scotland are brown.’ And the logician says, ‘No. There are cows in Scotland of which at least one is brown.’ And the mathematician says, ‘No. There is at least one cow in Scotland, of which one side appears to be brown.’ And this is funny because economists are not real scientists and because logicians think more clearly, but mathematicians are best.”

The Curious Incident of the Dog in the Night-Time (2003)