## tar.lrz

June 1, 2015

If presented with a tarball with a tar.lrz extension (e.g. after waiting many hours for one of these torrents to download), proceed as in the following example:

lrzuntar Ubuntu_14.04_LTS_sage-6.7-x86_64-Linux.tar.lrz

## The Prime Number Theorem (up to a constant factor)

April 14, 2015

It is suprisingly easy and pretty to give an elementary proof that the prime number theorem (that the number $\pi(x)$ of primes less than $x$ is asymptotic to $x/\log x$) is true up to a constant factor. (I will address only the part where we show that there are many (i.e. at least $~cx/\log x$) primes less than $x$.)

Consider a binomial coefficient ${n \choose k}$. Let $p$ be a prime. We first want to know what the largest power of $p$ is which divides ${n \choose k}$. There is more than one way to express the answer, the most elegant one I know of is as follows:

Add the numbers $k$ and $n-k$ in base $p$ (by the usual primary school algorithm). Then the number of carries that occur in this addition is equal to the greatest power of $p$ dividing ${n \choose k}$.

What is most important for us is the fact that this number is no more than $\log_p n$.

We immediately get

$\displaystyle \prod_{p\leq n}p^{\log_pn}\geq {n\choose k}$

which is of course equivalent to

$\displaystyle n^{\pi(n)}\geq {n\choose k}.$

Make a good choice of $k$ and we are within a constant factor of the prime number theorem!

## The Butterfly Theorem

March 1, 2015

Let $M$ be the midpoint of the chord $PQ$ of a circle. Let $AB$ and $CD$ be two other chords of the circle passing through $M$ and let $X$ and $Y$ be the intersection points of $PQ$ with $AD$ and $BC$ respectively. Then $M$ is the midpoint of the segment $XY$. Proof: The space of all degree two polynomials vanishing at the four points $A, B, C$ and $D$ is of dimension 2*. Thus it is spanned by the equation of the circle and the equation of the union of the lines $AB$ and $CD$. Choose coordinates such that the line $PQ$ is the $x$-axis and the point $M$ is the origin. Then the coefficient of $x$ in both our spanning polynomials is equal to zero, hence this coefficient is also zero in the equation of the union of $AD$ and $BC$. Therefore $M$ is the midpoint of the segment $XY$, as required.

*To see this there is a six dimensional space of degree two polynomials and we are imposing one linear condition each time we require the polynomial to vanish at a point. These four conditions are linearly independent since it is easy to find degree two curves passing through three of these points and not the fourth (e.g. a union of two lines).

February 11, 2015

Daniel discusses why what you see on your TV screen may be misleading. I wonder what the BCCI think of this.

An Oldie but a Goddie: Why South Africa has never won the f**ng World Cup.

On the Time Spent Preparing Grant Proposals. (hat-tip to SY).

The Dot and the Line (A Romance in Lower Mathematics) (youtube). This is actually a book, which I learnt on Saturday when I found the book in the Boundary Street Markets.

And last but not least, Australia will be competing in Eurovision 2015. And there is a petition calling on TISM to be our entry.

## The Quest for the Perfect Pad See Eiw

February 4, 2015

That is, the quest for the perfect Pad See Eiw that is not located in Sydney. This quest has now entered its tenth year and we now discuss last night’s entry, from Thai Nakonlanna in St Lucia, Qld.

Unfortunately there will be no pictures, even though I had my camera with me. Sorry if you’re the type of person that actually thinks a picture is worth a thousand words.

Before we get to the food, I want to mention the cost. The Pad See Eiw with chicken here was 12AUD. As a point of comparison, Thai La-Ong charges 10.5AUD (dinnertime price). There’s this myth floating around that Sydney is the most expensive city in Australia, but after seeing prices here in Brisbane I’m calling Bravo Sierra on that one.

With much anticipation the first mouthful left the plate, was delicately transported upwards on a fork and entered the mouth. In an instant, I knew that this was not the one. It’s hard for me to describe exactly what was wrong in words, though I feel it was not sweet enough.

I don’t know why I bother continuing on this futile quest. I really should give up.

## Ender’s Game

January 9, 2015

It was yonks and yonks ago that I read Orson Scott Card’s Ender’s Game and enjoyed the book so much that I happily continued with the sequels. It was with some trepidation however with which I approached the movie, as I had heard some indifferent reviews but also because of the following result:

Theorem: The book is always better than the movie.

(Proof: exercise left to the reader)

The movie follows exclusively the major plotline of the story. Thus it is essentially about Ender and Ender alone. It does at times skip through this storyline at a fast pace, sacrificing character development to do so which makes some of the scenes appear less meaningful or understandable when compared to the book version.

Disappointingly, the entire plot line involving Locke and Demosthenes is missing from the film. Thus if you’ve only seen the film, then you won’t get . The joke in is still understandable.

After watching the film, I felt I had to reread the book again to make a comparison between the two – as well as to test how accurate my memory of the book acually was. It felt weird reading dialogue which was inserted directly from the book into the film and thus I had just heard.

My recommendation is that the book is a must read. The movie is optional. Based on m y personal experiences with Jurassic Park and The Lost World (both of which I saw before I read), it is perhaps better to watch the movie before reading the book if you want to do both and for some reason haven’t yet read the book.

## On the Translation of Poetry

December 20, 2014

Taken at the Seoul Museum of Art.

## Flamingos!

October 5, 2014

Taken in Colombia.

## Sydney v Richmond

August 27, 2014

You are the captain of Sydney. The phone rings and you pick it up. Trent Cotchin is on the line.

“G’day Trent, what’s up?”

“I’m calling to discuss our game on this weekend. It seems that there’s an opportunity for us both to arrange things to benefit both of us.”

“Um, how?”

“Well what do you guys want to get out of this weekend’s match? The minor premiership, for sure, but even more importantly you want none of your players to get injured or suspended. Agree?”

“Yes”

“Well listen, I have a way to guarantee that you get all of the above. Interested?”

“Um, I’m not sure how you can guarantee that we don’t have any injuries.”

“All I ask for in return is that you let us win. Only by a small enough margin that you still keep top spot from the Hawks on percentage. But you see, we do really need the win more than you guys this time so we can make the eight.”

Trent pauses to take a breath, and continues

“Here’s how I propose we work things out. Come Saturday’s game, when the ball is bounced you allow us with no contest to score a couple of goals. Then both teams agree to sit down for the rest of the game and do nothing. We both get what we want out of the game without having to take any risks. Do you agree to this proposal?

(Also, if you accept, what will be the response of the rest of the world to such a highly contrived outcome?)

## Concyclicity in the Euclidean Plane

August 21, 2014

I want to address the question of how to determine whether or not four points are concyclic. In a previous post, this question reared its head and was solved by an introduction of the circumcentre of three of the points. In this post, I present a direct approach.

We prove the following:

Let $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$ be four points in $\mathbb{R}^2$. Then these four points are concyclic if and only if

$\displaystyle \left| \begin{array}{cccc} x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \\ x_4^2+y_4^2 & x_4 & y_4 & 1 \end{array} \right|=0.$

For the proof, first suppose that the points $(x_i,y_i)$ all lie on the curve

$\displaystyle (x-x_0)^2+(y-y_0)^2=r^2.$

The four equations we obtain give a linear dependence of the columns in the $4\times 4$ matrix under consideration, hence its determinant is zero.

Conversely, suppose that the determinant in question is zero. Then there is a nonzero vector in the nullspace of our $4\times 4$ matrix. If its first entry is nonzero, we can normalise it to be of the form $(1,-2x_0,-2y_0,x_0^2+y_0^2-r^2)^T$ and we obtain the circle that the four points lie on. If the first entry of this vector is equal to zero, then the four points must lie on a line, which after is all is just a degenerate circle, QED.