Apparently, it's not Debian planet that has problems to deal with MathJax (which makes posts usingn it appear illegible on Debian planet), but the ikiwiki plugin that I am using which generates garbage in the feeds that get consumed by

If you it know of a better plugin (which doesn't generate such output) please let me know, especially if it is not super computationally expensive for a Core 2 Duo T7250 (which is my notebook).

The real thing

As William Stein is now offering a course on Number Theory and he has been posting recorded videos of his lectures, I started watching some of them (mainly, the ones regarding continued factions). In particular, he shows the usual recursive formula for convergents of a continued fraction and that's super nice.

For the curious reader, the recurrence relations for the convergents of \([a_0, a_1, \ldots, a_n, \ldots]\) are: \[ p_n = a_n p_{n-1} + p_{n-2}\\ q_n = a_n q_{n-1} + q_{n-2}, \] with initial conditions (\(p_{-2} = 0\), \(p_{-1} = 1\), \(p_0 = a_0\), \(q_{-2} = 1\), \(q_{-1} = 0\), \(q_0 = 1\)).

He even motivated the use of continued fractions with the golden ratio, which is super nice, given that I like the subject and have been writing a document collecting facts that I know about the Fibonacci numbers (well, this document is horribly incomplete and not even close to something that I would consider proper for public consumption—I plan on publishing them soon).

OK, after he discussed the basics of it convergents, he noted that the recurrence relations are like those defining the Fibonacci numbers, just that one of the terms is "weighted" by the coefficients of the continued fraction.

I missed one thing, though: neither his book nor wikipedia's article on continued fractions mentions a very neat, alternative way to express the convergents of continued fractions (truth be said, I took a quick peek at wikipedia's article and I found that it doesn't mentionat least explicitly, in a 1 min skim—it may well be buried somewhere else).

I, then, proposed the following exercise for his students (which he apparently liked, as he +1's the suggestion):

Prove that the recurrence relation of the \(p_i\)'s and \(q_i\)'s that we mentioned before can be obtained via matrix multiplication. More precisely, prove that: \[ \begin{bmatrix} a_0 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_1 & 1\\ 1 & 0 \end{bmatrix} \cdots \begin{bmatrix} a_n & 1\\ 1 & 0 \end{bmatrix} = \begin{bmatrix} p_n & p_{n-1}\\ q_n & q_{n-1} \end{bmatrix}. \] As a corollary, derive Cassini's identity for the Fibonacci Numbers.

Please, if you any errors on this, please let me know so that I can fix it.

Edit: Thanks "noone" for spotting a typographical error.

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