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Euler, pirates, and the discovery of America

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Is there nothing Euler wasn’t involved in?!

That rhetorical question is independent of the following two, which are exceedingly weak connections.

“Connections” to piracy: Very tenuous connections, of course, but briefly, summarising from the article:

  1. Maupertuis: President of the Berlin Academy for much of the time Euler was there. His father got a license from the French king to attack English ships, made a fortune, and retired. Maupertuis is known for formulating the Principle of Least Action (but maybe it was Euler), and best known for taking measurements showing the Earth bulges at the equator as Newton had predicted, thus “The Man Who Flattened the Earth”.
  2. Henry Watson: English privateer living in India, lost a fortune to the scheming British East India Company. Wanted to be a pirate, but wasn’t actually one. Known for: translated Euler’s Théorie complette [E426] from its original French: A complete theory of the construction and properties of vessels: with practical conclusions for the management of ships, made easy to navigators. (Yes, Euler wrote that.)
  3. Kenelm Digby: Not connected to Euler actually, just the recipient of a letter by Fermat in which a problem that was later solved by Euler was discussed. Distinguished alchemist, one of the founders of the Royal Society, did some pirating (once) and was knighted for it.
  4. Another guy, nevermind.

Moral: The fundamental interconnectedness of all things. Or, connections don’t mean a thing.

The discovery of America: Columbus never set foot on the mainland of America, and died thinking he had found a shorter route to India and China, not whole new continents that were in the way. The question remained whether these new lands were part of Asia (thus, “Very Far East”) or not. The czar of Russia (centuries later) sent Bering to determine the bounds of Russia, and the Bering Strait separating the two continents was discovered and reported back: America was not part of Russia. At about this time, there were riots in Russia, there was nobody to make the announcement, and “Making the announcement fell to Leonhard Euler, still the preeminent member of the St. Petersburg Academy, and really the only member who was still taking his responsibilities seriously.” As the man in charge of drawing the geography of Russia, Euler knew a little, and wrote a letter to Wetstein, member of the Royal Society in London. So it was only through Euler that the world knew that the America that was discovered was new. This letter [E107], with others, is about the only work of Euler in English. That Euler knew English (surprisingly!) is otherwise evident from the fact that he translated and “annotated” a book on ballistics by the Englishman Benjamin Robins. The original was 150 pages long; with Euler’s comments added, it was 720. [E77, translated back into English as New principles of gunnery.]

Most or all of the above is from Ed Sandifer’s monthly column How Euler Did It.

The works of Leonhard Euler online has pages for all 866 of his works; 132 of them are available in English, including the translations from the Latin posted by graduate student Jordan Bell on the arXiv. They are very readable.

This includes his Letters to a German Princess on various topics in physics and philosophy [E343,E344,E417], which were bestsellers when reprinted as science books for a general audience. It includes his textbook, Elements of Algebra [E387,E388]. Find others on Google Books. The translations do not seem to include (among his other books) his classic textbook Introductio in analysin infinitorum [E101,E102, “the foremost textbook of modern times”], though there are French and German translations available.

Apparently, Euler’s Latin is (relatively) not too hard to follow.

Written by S

Mon, 2010-03-01 at 23:04:23

Lattice points visible from the origin

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[A test of LaTeX-to-Wordpress conversion. Bugs remain, point them out. Original PDF]

Here is a problem I love. It is simple to state, and it has a solution that is not trivial, but is easy to understand. The solution also goes through some beautiful parts, so I can promise it’s worth reading :-)

[The solution is not mine. Also, the question is just an excuse for the ideas in the solution. :P]

Question. Suppose you are standing on an infinite grid in the plane. You can see infinitely in all directions, but you cannot see through grid points: a point is hidden from view if some other grid point lies in your line of sight. What fraction of the grid points can you see?

Let us first imagine that we are standing at the origin, and that the grid is that of the lattice (integer) points.

The blue points are visible; the grey points are not

The blue points are visible; the grey points are not

Read the rest of this entry »

Written by S

Fri, 2008-11-07 at 15:00:30

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A theorem by Euler on partitions

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There are so many beautiful facts in mathematics that are not as well known as they deserve to be. [Just today I couldn’t find an online reference for the Robinson-Schensted(-Knuth) correspondence.] Some efforts in the direction of communicating these to an online audience have been collected in the Carnival of Mathematics (41 so far) posts on various blogs. Another good idea I recently found is Theorem of the Day. [EtA: Found at Theorem of the Day: RSK correspondence.]

Anyway… today I attended a lecture on Euler by William Dunham (great talk!) where after a brief biography of Euler and a tour through some of his results, he showed Euler’s proof of a particular theorem about partitions. It will be familiar to anyone who has read about integer partitions, but it deserves to be familiar to everyone! So here is a sketch of the proof, for your reading pleasure. [It doesn’t do justice to present only a “sketch”, but I have assignments due tomorrowtoday… feel free to take it and turn it into some form worthy of the content. :-)]

Now when I say “a theorem by Euler”, it is very ambiguous which one I mean, and even with the “on partitions” restriction, there are several. So let me state it:

The number of partitions of a number into odd parts is the same as the number of partitions of the number into distinct parts.

[If I remember correctly what was narrated in the talk, one of the Bernoullis wrote to Euler asking him a question about partitions, and within a few days (think of the postal system then) Euler replied with a proof of the theorem, along with a apology for the delay caused by the poor eyesight he had recently been suffering from. This is an outline of his proof, ask me about the details.]

Let D(n) denote the number of partitions into distinct parts, and O(n) denote the number of partitions into odd parts. Then we have:
\displaystyle \sum_{n\ge0}D(n)x^n
\displaystyle = (1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)\dots
\displaystyle = \frac{1-x^2}{1-x}\frac{1-x^4}{1-x^2}\frac{1-x^6}{1-x^3}\frac{1-x^8}{1-x^4}\frac{1-x^{10}}{1-x^5}\dots
\displaystyle = \frac{1}{(1-x)(1-x^3)(1-x^5)\dots}
\displaystyle = (1+x+x^{1+1}+\dots)(1+x^3+x^{3+3}+\dots)(1+x^5+x^{5+5}+\dots)
\displaystyle = \sum_{n\ge0}O(n)x^n
which proves the theorem.
The first equality, which gives the generating function for the D(n)s, can be seen as follows: when you “expand” and write out (1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)\dots, you get 1 + x + x^2 + (x^3+xx^2) + (x^4+xx^3) + (x^5+xx^4+x^2x^3) + \dots, where the coefficient of any term x^n is exactly the number of ways of writing x^n as a product of distinct factors of the form x^k, which is D(n). Similarly, the last equality, about the generating function for the O(n)s, is because the coefficient of any term x^n in (1+x+x^{1+1}+\dots)(1+x^3+x^{3+3}+\dots)(1+x^5+x^{5+5}+\dots) is exactly the number of ways of writing n as a product of (not necessarily distinct) x^k for odd k.

The result might not be terribly important, but this idea was the starting point for proving several observations about partitions, and the method has led to the discovery of several facts and is more widely applicable than you think!

If you want a combinatorial proof by bijection (“For Entertainment Only”) here is an outline of one:
Given a partition of n into odd parts, say
n = a_11 + a_33 + a_55 + \dots,
write each a_i “in binary”, by which I mean as a sum of distinct powers of two. So you have
n =  (2^{b_{11}}+2^{b_{12}}+...)1 + (2^{b_{31}}+2^{b_{32}}+...)3 + \dots.
Now just get rid of the brackets, and note that all terms are distinct.
E.g. for 19 = 5+3+3+3+1+1+1+1+1,
i.e. 19 = (1)5 + (2+1)3 + (4+1)1, you get the partition
19 = 5 + 6 + 3 + 4 + 1.
[Exercise: Why are all parts distinct?]

Conversely, given a partition into distinct parts, we can separate the “even part” from the “odd part” in each term, so for example with 19 = 5+6+3+4+1, we write
19 = 5 + (2)3 + 3 + (4)1 + 1, and collect the coefficients of each odd number, so 19= 5 + (2+1)3 + (4+1)1, which was our original odd partition.
[Exercise: Why can’t we do the same thing starting with any partition?]

[Question: Is the bijective proof related to the algebraic one?]

Written by S

Wed, 2008-10-15 at 06:26:39

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