March 2018 – The Melancholy Roman

10 March 20184 August 2019

The best opening paragraphs in non-fiction

Some famous novels have famous opening sentences. There’s Anna Karenina, for example:

All happy families are alike; each unhappy family is unhappy in its own way.

Or A Tale of Two Cities:

It was the best of times, it was the worst of times…

though to be strictly accurate, that one keeps going for quite a bit longer.

There’s Moby Dick:

Call me Ishmael.

There’s Rebecca, which I’ve never read, though I did see the Hitchcock film. The opening sentence is:

Last night I dreamt I went to Manderley again.

And there are famous sentences that feel instinctively like opening lines, but aren’t. My go-to example for this is from To Kill a Mockingbird:

Maycomb was an old town, but it was a tired old town when I first knew it.

That one actually comes a couple of pages in. The real opening sentence is the rather mundane: “When he was nearly thirteen my brother Jem got his arm badly broken at the elbow.”

All these are examples from fiction. Non-fiction is rather more restricted in being limited to the facts and so there are no classic opening sentences that I know of.

But there are some great opening paragraphs. The Man on Your Conscience is a mostly forgotten book by a man named Michael Eddowes. It’s about the notorious Rillington Place case in Britain in the 1950s. Timothy Evans was arrested, tried, convicted and hanged for the murders of his wife and daughter. A few years later it was discovered that his landlord was a serial killer and that Evans was probably innocent.

Eddowes’ book isn’t nearly as famous as Ludovic Kennedy’s work about the same case, 10 Rillington Place. People who’ve read both agree that Kennedy was the better writer and he was certainly a prominent journalist and decades-long campaigner against miscarriages of justice. But I suspect that even Kennedy was impressed by Eddowes’ superb and factually accurate opening paragraph:

The leading players were soon to die. The judge was dead within a few weeks, the accused man was hanged within a few weeks, within three years the Crown’s second witness was strangled by the Crown’s chief witness, and the Crown’s chief witness was hanged.

For a few years I thought that that was the best opening paragraph in all of non-fiction. But then a friend recommended I read Edward Gibbons’ The History of the Decline and Fall of the Roman Empire.

Gibbons’ work is a masterpiece, still famous centuries after he finished writing it. It’s also a great read and the opening paragraph is this:

In the second century of the Christian era, the empire of Rome comprehended the fairest part of the earth, and the most civilised portion of mankind. The frontiers of that extensive monarchy were guarded by ancient renown and disciplined valour. The gentle, but powerful, influence of laws and manners had gradually cemented the union of the provinces. Their peaceful inhabitants enjoyed and abused the advantages of wealth and luxury. The image of a free constitution was preserved with decent reverence. The Roman senate appeared to possess the sovereign authority, and devolved on the emperors all the executive powers of government. During a happy period of more than fourscore years, the public administration was conducted by the virtue and abilities of Nerva, Trajan, Hadrian, and the two Antonines. It is the design of this and of the two succeeding chapters, to describe the prosperous condition of their empire; and afterwards, from the death of Marcus Antoninus, to deduce the most important circumstances of its decline and fall: a revolution which will ever be remembered, and is still felt by the nations of the earth.

Now that’s what I call an opening paragraph!

The Melancholy Roman

7 March 201810 November 2018

The Binomial Theorem

This picture has absolutely nothing to do with this blog post.

Or: practicing using LaTeX in WordPress.

We’ll start with factorial notation. In mathematics, $n!$ reads “ $n$ factorial” and signifies $n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1$ . For example,

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120.$

Now, imagine that we have $n$ objects and we want to select $k$ of them (assuming that $k \leq n$ ). How many different ways could we do this?

Assuming we can’t count the same object twice, we have $n$ choices for the first object, $n-1$ choices for the second object, $n-2$ choices for the third object and so on. In fact we have

$\underbrace{ n \times (n-1) \times \ldots \times (n-k+1)}_{k\ \text{terms}} \ \ \text{choices}.$

We can put this much more simply as

$\frac{n!}{(n-k)!}.$

If you can’t quite see this, consider that

$n! = n \times (n-1) \times \ldots \times (n-k+1) \times (n-k)!$

and then look at the expansion below and see how the terms cancel out:

$\frac{n!}{(n-k)!} = \frac{ n \times (n-1) \times \ldots \times (n-k+1) \times \text{\sout{$(n-k)!$}} }{ \text{\sout{$(n-k)!$}} }.$

So, to be clear, if we have $n$ objects and we are choosing any $k$ of them,

$\frac{n!}{(n-k)!}$

gives the total number of possible permutations.

But what if we don’t care about permutations? What if, selecting $k$ different objects, we don’t care what order they’re in? The above formula will give many duplications. For example, if we’re picking three different letters from $A$ , $B$ , $C$ , $D$ , $E$ , $F$ and we choose $A$ , $B$ and $E$ , there are six different ways of doing this: $ABE$ , $AEB$ , $EAB$ , $EBA$ , $BEA$ , $BAE$ . But they’re all equivalent, just different permutations of the same three letters.

How many such permutations are there? If you’re arranging $k$ different objects, there are $k$ choices for the first item, $k-1$ choices for the second item, $k-2$ for the third and so on. In fact there are $k \times (k-1) \times (k-2) \times \ldots \times 1 = k!$ possible permutations.

Now, since

$\frac{n!}{(n-k)!}$

counts every possible permutation, and since $k$ objects must have $k!$ possible permutations, we simply divide by $k!$ to give the number of ways of choosing $k$ objects from $n$ if we don’t care about the order. This gives us

$\frac{n!}{k!(n-k)!}.$

Next we introduce some notation. The symbol

$\binom{n}{k}$

is pronounced “ $n$ choose $k$ ” and means that given $n$ objects, we want to choose $k$ of them. This notation is used when we don’t care about permutations and as you may have guessed, we define

$\binom{n}{k} = \frac{n!}{k!(n-k)!}.$

To demonstrate how useful this is we’ll give an example. Imagine we have $5$ coins on a table and we want to pick up any $2$ of them. How many ways can we do this?

A bit of mental visualisation suggests that the answer is $10$ but we can apply the formula:

$\binom{5}{2} = \frac{ 5! }{ 2! (5-2)! } = \frac{ 5 \times 4 \times 3 \times 2 \times 1 } { (2 \times 1 ) \times (3 \times 2 \times 1) } = 10.$

Clearly more complicated examples can be beyond the power of mental visualisation and then the formula comes into its own.

So much for the introduction. We can now get to the nitty gritty. Let $a, b \in {\mathbb{C}}$ . Then

$(a+b)^0 = 0 \ \ \text{ (of course*)}$

and

$(a+b)^1 = a+b,$

and

$\begin{align*} (a+b)^2 &= (a+b)(a+b)\\ &= a^2 + 2ab + b^2, \end{align*}$

and indeed

$\begin{align*} (a+b)^3 &= (a+b)(a+b)(a+b)\\ &= a^3 + 3a^2b + 3ab^2 + b^3. \end{align*}$

These calculations are fairly easy to do by hand but as $n$ gets bigger the algebra for calculating $(a+b)^n$ rapidly becomes tedious. We know that the terms will be $a^n$ , $a^{n-1}b$ , $a^{n-2}b^2$ and so on, but what are the coefficients?

Every term in the multiplied out expansion is the product of an $a$ or a $b$ in each $(a+b)$ term. And this means we can easily work them out.

For instance, let us take the $a^3b^7$ term in the expansion of $(a+b)^{10}$ . We choose the $a$ term from three lots of $(a+b)$ and the $b$ term from seven lots of $(a+b)$ . So how many times would $a^3b^7$ occur if the entire product was multiplied out? The answer is, as many times as there are ways of choosing $3$ of the bracketed terms to give us the $a$ or $7$ of them to give us the $b$ . That is, $\binom{10}{3} = \binom{10}{7} = 120$ .

Generalising, if we multiply out $(a+b)^n$ , each $a^{n-k}b^k$ term occurs $\binom{n}{n-k}$ times. Since $\binom{n}{n-k} = \binom{n}{k}$ and the latter is simpler to write, we say that the coefficient for $a^{n-k}b^k$ is $\binom{n}{k}$ .

And now we can state the compact version of the Binomial Theorem. By convention, we define $0! =1$ . For $a, b \in {\mathbb{C}}$ and $n \in {\mathbb{N}}$ , the Binomial Theorem says that

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k.$

As ever, we can write the expanded version, which begins

$(a+b)^n = \binom{n}{0}a^nb^0 + \binom{n}{1}a^{n-1}b^1 + \ldots$

Postscript*

As generations of mathematics students have reasoned, if (for example)

$1^0 = 5^0 = 496^0 = x^0 = p^0 = 1,$

that is, if anything to the power of zero is one; while (for example)

$0^1 = 0^{5} = 0^{496} = 0^x = 0^p = 0,$

that is, zero to the power of anything is zero; then what is the solution to

$0^0?$

The answer is that we generally say it is undefined. However, in the context of the Binomial Theorem, by convention we define $0^0 = 1$ in order to make the equation work.

Second Postscript

The caption under the image at the top of this blog post says it has absolutely nothing to do with the rest of the post. That is correct. It is actually detail of a miniature painting called Carretó fantasma by Salvador Dalí, and is displayed in the Dalí Museum in Figueres, Spain.

The Melancholy Roman

3 March 20184 August 2019

Found in change

I mentioned in my previous post that Australia used to have British-style pounds, shillings and pence. When the currency was decimalised on 14 February 1966 they were replaced with dollars and cents. For convenience the new coin sizes were copied directly from the old.

All this I knew in an academic kind of way but I’d never seen evidence of it till today. Looking in my change, I noticed that one of my 20c coins had a design I’d never seen before.

That wasn’t too unusual. The Australian Mint does commemorative coins and I’d been overseas for most of the past 16 years. It was easily possible that there were coin designs I didn’t know about.

But it did seem unusual that it would be a 20c, so I looked a bit closer. And I realised it wasn’t a 20c at all.

I still doubted myself and thought that maybe it was a facsimile – that is, a commemorative coin marking the currency that Australia used to have. But I turned it over and saw an unfamiliar head and the inscription Georgivs V. It was this detail that convinced me, since anything commemorative would still have the Queen’s features on it. This one had a man and I concluded that it was genuinely old. Georgius V was probably George V, the Queen’s grandfather.

You can see the picture of both sides of it, above. It’s a two shilling coin from 1931. How I got it in change I don’t know. Maybe someone knew they had it but decided to get rid of it. Perhaps someone accidentally mixed it with their spare change and is now cursing themselves; maybe it has freakishly remained in circulation for the past 52 years. Certainly I haven’t had it more than a couple of days.

But is it worth anything? My guess is, about 20c. If it was in mint condition it might be collectible and moderately valuable. But let’s face it, it’s a battered old coin. I’ll retain it as an interesting keepsake but that’s definitely where its value lies.

The Melancholy Roman