Talk:Continued fraction

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Field: Analysis

If zero is allowed

The article on e (mathematical constant) says, as this article says, that the CF is

$e = [[2; 1, \textbf{2}, 1, 1, \textbf{4}, 1, 1, \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots, \textbf{2n}, 1, 1, \ldots]], \,$

But it adds that it can be written more harmoniously by allowing zero: (Hofstadter, D. R. "Fluid Concepts and Creative Analogies: Computer Models of the Fundamental Mechanisms of Thought" Basic Books, 1995)

$e = [[1 , \textbf{0} , 1 , 1, \textbf{2}, 1, 1, \textbf{4}, 1 , 1 , \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots]]. \,$

A scan of this article shows me no explanation to help me interpret that second expression. — CpiralCpiral 00:50, 30 November 2011 (UTC)

It seems that they used a comma rather than the normal initial semicolon to make the pattern more visually obvious. They are saying that if the rules on continued fractions are relaxed to allow zero, then the two expressions are equivalent (the second being more "hamonious" because with the zero there is an obvious pattern from the start). The expression with the zero is simply
$1 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{4 + \ddots}}}}}}}$
The Continued fraction article seems to give enough to interpret the expression. It does state: "Note that it is customary to replace only the first comma by a semicolon", which should be enough of a hint. Quondumtalkcontr 04:11, 30 November 2011 (UTC)
So you're saying that there they did two steps to get from one version to the other
• [1;0,1...] is eq to [2;...] and
• "The semicolon in the square and angle bracket notations is sometimes replaced by a comma"
And that the "harmony" is thereby two things
• the pattern of numbers
• the consistent use of commas
Now is there mention here of a rule (perhaps buried in the convergents formulas or the theorems) that a zero can relax, for any reason, as any but the leftmost quotent? It says "first integer may be zero", but that's it. The articles should harmonize. — CpiralCpiral 08:57, 30 November 2011 (UTC)
It does say "... where $a$0 is an integer, any other $a_i$ members are positive integers" under the section Basic formulae. I've edited the article to make it clear that this also applies to the infinite case. So the rule about all but the first value being positive (and hence non-zero) is there. The relaxation to zeros falls under the section Generalized continued fraction (perhaps not very obvious). The article unfortunately does not have a clearcut Definition section. Quondumtalkcontr 09:40, 30 November 2011 (UTC)
• On the contrary, it seems that, since the section Generalized continued fraction follows the section Some useful theorems, which say "positive integers", that a generalized fraction must not allow zero.
• Does the generalized fraction have a bracket notation? The main article has none.
Join the discussion at Talk:E (mathematical constant)#Continued fraction notation? — CpiralCpiral 10:23, 7 December 2011 (UTC)
For clarity, no restriction applies to the values used in a "generalized continued fraction"; they may be zero. The concise notation is not adequate for the purpose; aside from that I don't know. Quondumtalkcontr 12:21, 7 December 2011 (UTC)

Theorem 5

I think there is some mistake in theorem 5 - what I mean is if $\alpha =[a_0, a_1, \cdots, a_n, a_{n+1}, \cdots]$, then $\alpha = \frac{\alpha_{n+1}h_n+h_{n-1}}{\alpha_{n+1}k_n+k_{n-1}}$, where $\alpha_{n+1} =[a_{n+1}, \cdots]$, so:
$\left|\alpha - \frac{h_n}{k_n}\right|= \frac{1}{(\alpha_{n+1}k_n+k_{n-1})k_n} < \frac{1}{(k_n+k_{n-1})k_n}$

since $\alpha_{n+1} > 1$ strictly for infinitive fractions. What do you think? — Preceding unsigned comment added by 82.41.56.99 (talk) 00:56, 4 January 2012 (UTC)

Try α=√2. The convergents are 3/2, 7/5, 17/12, ... . |3/2-√2|=1/11.65 which is between 1/10 and 1/14, |7/5-√2|=1/70.35 which is between 1/60 and 1/85, and |17/12-√2|=1/407.64 which is between 1/348 and 1/492, so while I'm not sure where your logic went wrong the theorem appears to hold.--RDBury (talk) 14:30, 4 January 2012 (UTC)

Introduction

The introduction starts off by taking about the continued fraction associated to a number. Would it be better to define a continued fraction intrinsically, say that every CF represents (evaluates to/converges to) a real number and then sy that every real number has a CF? Thue Siegel Roth (talk) 21:11, 8 February 2012 (UTC)

It's probably a matter of taste but I like it better the way it is. The Generalized continued fraction article would probably be the best place to cover convergence issues. The way I thinking about it is that, for example, a decimal expansion is really a infinite series, but we don't (I hope) start our article on the decimal system with material on convergence tests.--RDBury (talk) 06:30, 9 February 2012 (UTC)
I don't copmment much on Wikipedia, but wanted to say that I thought the example was great. It would probably be more encyclopedic to present the general case, but using a specific example, at least for me, was much more edifying. Good job.Zipperfish (talk) 16:12, 15 March 2013 (UTC)

Merge discussion

The article Convergent (continued fraction) does not discuss anything that would not be better discussed in context here. I propose merging it into this article. RockMagnetist (talk) 17:48, 14 January 2013 (UTC)

As a separate article it does not have any merit, since convergents are discussed here in all details. Also, I doubt that the dab page Convergent is a good solution, but still unsure whether it should be redirected to Limit (mathematics) #Convergence and fixed point. Incnis Mrsi (talk) 18:34, 14 January 2013 (UTC)