Jonathan David Page talks about whatever he happens to be thinking about. Sometimes other people join in.

Regarding to the "posting links regularly" thing I mentioned in the last post, I need to make a remark:

HAHAHAHAHAHAHAHAHAHAHAHAHAHA

hahahaha

wheezehahaha. ha.

Okay, now that that's over with, I can get down to business.

No, it isn’t Pi Day, or anything resembling it. I was editing some stuff, and I noticed the original “Approximating Pi” article. I’d been meaning to rewrite the code since I learned about continued fractions, and since I didn’t have anything better to do, This is a lie. I do, in fact, have a number of things to do that various people might define as “better”, but I didn’t feel like doing any of them. I felt like sitting around in my bathrobe and writing Python. I decided to do so. A detailed explanation follows the source code, since this is less mathematically facile than the previous version.

You didn't get a post for July 22, so here's some Python which finds approximations for pi.

Update 20 Dec 2012: See also the revised version utilizing continued fractions.

In honor of everyone's favourite mathematical constant, March 14, or 3/14 is Pi Day. Hooray!

Next up: Tau Day on June 28th, or 6/28. After that we get "European Pi Approximation Day" on July 22nd, or 22/7. Such fun we can look forward to!

Ever since I first ran into this proof a few years ago, I've always thought it rather elegant. It came up (read: I brought it up) while studying for a proofs test, so I thought I would post it here. It serves as a decent test of whether the mathematics typesetting system on here is working properly.

\(\sqrt 2\) is irrational.

Suppose \(\sqrt 2\) is a rational number. This means that it can be expressed as an irreducible fraction \(\frac{p}{q}\) where both \(p\) and \(q\) are integers. So \(\sqrt{2} = \frac{p}{q}\), thus \(2 = \left(\frac{p}{q}\right)^2\), thus \(2 = \frac{p^2}{q^2}\), so \(2q^2 = p^2\). Since \(p^2\) has a factor of 2, \(p^2\) is even, and therefore \(p\) is also even, meaning that \(p = 2k\) where \(k\) is an integer. Substituting, \(2q^2 = (2k)^2\), thus \(2q^2 = 4k^2\), thus \(q^2 = 2k^2\). Since \(q^2\) has a factor of 2, it is also even, so \(q\) is even. Therefore, \(p\) and \(q\) are both even, and so share a common factor of 2. But we assumed that \(\frac{p}{q}\) was irreducible. This is a contradiction. Therefore \(\sqrt 2\) is irrational. □

The Ackermann function is a mathematical function that returns really really large numbers. For example, \(A(4,2)\) is an integer with over 19,700 decimal digits.

Graham's number is a holy-moly massively horrifyingly huge number which makes a googolplex look like diddly squit. Actually, it makes numbers which make numbers which make googolplex look like diddly squit look like ant droppings look vanishingly small. It it unimaginably huge. I do not have words to explain how much massively huger than pretty much any number you can think of it is.

As you can imagine, \(A(g_{64}, g_{64})\) is quite a large number.

PS-- I'll answer the \(e^{\pi i}\) question soon, I want to type out a proper explanation for that.