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Q: What is this notation $[b_0 b_1,...]_\alpha$ with continued fractions in the image below? Especially what does subscript $]_\alpha$ mean?

enter image description here

What I have tried: I know the notation $$[b_0 b_1,...]=b_0+\frac{1}{b_1+}\frac{1}{b_2+\ldots }$$ for a simple cfrac. I tried to google translate the text but couldn't make any sense of it. So after translatng the text

Lause 4.11.Olkoon훼∈ℝ∖ℚ, 훼 >0annettu ja olkoon[푏0;푏1,...]훼(4.109)Ketjumurtoalgoritmilla muodostettu lukuun훼liittyvä yksinkertainen ketju-murtolukukehitelmä. Tällöin

I get

Theorem 4.11.Let 훼 ∈ℝ ∖ ℚ, 훼> 0 be given and let [푏 0; 푏 1, ...] 훼 (4.109) A simple chain-fractional development related to the number 훼 formed by a chain fracture algorithm. Here

I couldn't understand what this means. I hope someone who knows Finnish language can answer this question, or anybody who has encountered the notation. I haven't seen this kind of notation anywhere else. Thanks.

Peppa Pig
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  • Isn't the notation with index $\alpha$ defined earlier? – md2perpe Aug 28 '20 at 19:42
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    Native Finnish speaker here (but it's past me bed time). Can you add a link to the source, please. That would probably clear up the missing pieces, the notation $]_\alpha$, in particula. Google translate gave its best shot, but it is not trained in mathematical jargon. Undoubtedly you guessed that ketju=chain and murtoluku=fraction, but ketjumurtoluku=continued fraction. Kehitelmä could mean development in some other context, but here it means expansion. Finnish uses kehitelmä also in the context of partial fraction expansion (of a rational function), and – Jyrki Lahtonen Aug 28 '20 at 22:01
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    (cont'd) my guess is that this was recycled by whoever first introduced continued fraction -terminology into Finnish. It seems clear to me that the notation $[b_0,b_1,\ldots,b_k]_\alpha$ was introduced and explained earlier, and it refers to the output of a prescribed method for producing the continued fraction expansion of a real number $\alpha$. There's no reason to think it would not be the usual algorithm. – Jyrki Lahtonen Aug 28 '20 at 22:06
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    Our old boys were brought up with German math jargon, where also Kette=chain/necklace, Kettenbrüche=continued fractions. This is a very plausible theory. We translated a lot of terminology from German back in the day. – Jyrki Lahtonen Aug 28 '20 at 22:09

1 Answers1

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I'm just going to do my best to make a translation from the Google-English:

Thm 4.11: Let $\alpha \in R \setminus Q, \alpha > 0$ be given, and let $$ [b_0, b_1, \ldots]_\alpha $$ be the simple continued-fraction expression produced by the continued-fraction algorithm. Then $$ \alpha = [b_0, b_1, \ldots]_\alpha. $$

In short: the continued fraction algorithm (presumably described earlier in the paper) produces an expression which, when evaluated, gives $\alpha$. More precisely, the partial terms of the CF form a sequence which has a limit that is $\alpha$.

In really short: the continued fraction algorithm works to produce continued fractions as expected.

The proof(?) that follows appears to say 'suppose that you had two different cfracs for $\alpha$, and that the first difference is in the $m$th term. Then (by various bits of algebra) you'd find (some algebraic relation between the two differing $m$th terms).

John Hughes
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  • Thanks. I need some time to process this. – Peppa Pig Aug 28 '20 at 18:52
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    @PeppaPig The most important part is for you to practice with actual continued fractions. You can click on the continued fractions tag at your answer. The abstract definition of them is alphabet soup, but actually calculating them with an ordinary calculator and producing the first several "convergents" is really easy... – Will Jagy Aug 28 '20 at 21:02
  • $$ \begin{array}{cccccccccccccccccccccccc} & & 3 & & 1 & & 1 & & 1 & & 1 & & 6 & & 1 & & 1 & & 1 & & 1 & & 6 & \ \ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 3 }{ 1 } & & \frac{ 4 }{ 1 } & & \frac{ 7 }{ 2 } & & \frac{ 11 }{ 3 } & & \frac{ 18 }{ 5 } & & \frac{ 119 }{ 33 } & & \frac{ 137 }{ 38 } & & \frac{ 256 }{ 71 } & & \frac{ 393 }{ 109 } & & \frac{ 649 }{ 180 } \ \ & 1 & & -4 & & 3 & & -3 & & 4 & & -1 & & 4 & & -3 & & 3 & & -4 & & 1 \end{array} $$ This is for $\sqrt {13},$ below $\frac{p}{q}$ is $p^2 - 13 q^2$ – Will Jagy Aug 28 '20 at 21:03
  • +1 I think this is pretty much all that's going on. – Jyrki Lahtonen Aug 28 '20 at 22:11
  • Excellent. Is there a "Finnish" badge? :) – John Hughes Aug 28 '20 at 23:29
  • Let's see if the OP waves the Fin(n)ish flag to honor you being the first to cross the line. – Jyrki Lahtonen Aug 29 '20 at 05:07