5 ms·
This is a lovely and gentle (you hardly realize it) to a lot of very deep mathematics... great post, thanks to the author! There's a lot I learned and would lov
by svat 8y ago
This is a lovely and gentle (you hardly realize it) to a lot of very deep mathematics... great post, thanks to the author! There's a lot I learned and would love to look up and continue to explore.
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As an aside, one thing I like to point out though is that the definition of “good approximation” seems to some extent determined by what has the cleanest theory, than what one may naively desire, as in this paragraph from the article:
> Emily consider ways of giving each answer a score. Initially, she thought that for each fraction, the score could be the (absolute) difference between her number and the proposed fraction, and then multiplied by the denominator. (The lower the better). However, after talking to some of her tech friends, she decided to make it even stricter [...] denominator squared.
A similar thing comes up in many expositions of “best rational approximation” in books and on the internet, where instead of |x-p/q| we use |q(x-p/q)| = |qx-p|, and here in this post for even cleaner theory we're using |q(qx-p)|. A post I wrote a while ago to clarify this issue, with a small C program: https://shreevatsa.wordpress.com/2011/01/10/not-all-best-rational-approximations-are-the-convergents-of-the-continued-fraction/ https://shreevatsa.wordpress.com/2011/01/10/not-all-best-rat...
- extremelearning 8y agoI pondered this issue for what seemed like an inordinate amount of time namely: on how to describe this subtle but key difference. Unfortunately, I couldn't find a nice way, so i glossed over this point, which you correctly say makes many expressions and theorems cleaner and more elegant. Furthermore, this difference helps explains why q^2 is a natural choice, which some other readers on this thread have enquired about.
- ComplexSystems 8y agoHow do the results change if you use multiplicative error rather than additive error? That is, rather than |x-p/q|, you use max(x/(p/q), (p/q)/x). This is sometimes useful when trying to approximate rationals.
- extremelearning 8y agoGood question, but I don't know and haven't really done anything substantially related that might even give us a hint. Hopefully someone else chime in on this thread. ;)