Talk:Continued fraction/Archive 1

message: A few comments.
The definition of continued fractions here is too narrow. First of all, continued fraction theory is divided into analytic and arithmetic theory. The entry here is devoted entirely to arithmetic theory. In analytic theory, which is the focus of about half of the research, the numerators are not confined to being one. In fact, it is best to consider a continued fraction to be a formal expression of indeterminates (with numberator and denominator indeterminates.) For an infinite continue fraction, the indeterminates are taken in a complete normed field so that convergence can be defined. Continued fractions with elements from p-adic fields, formal Laurent series fields etc, have all been considered. These all fall outside of even the analytic and arithmetic categories I have mentioned above. With just formal indeterminates, continued fractions are even studied combinatorially. See the work of Flajolet, for example. BTW, for information purposes, I will mention that analytic theory refers to complex analysis, that is, calculus in the complex plane. So in this HUGE area of research, the elements of the continued fractions are assumed to be complex numbers. (Indeed, continued fractions are often viewed as products of fractional linear transformations, which are beautifully understood in the context of the complex plane.)
The associated entry on "generalized continued fractions" is even more naive. The expressions there are what are generally refered to as continued fractions, only the elements of the expression are not confined to being integers; indeed even in some of the examples given, for instance for exp(x), x is given as being complex, contraticting the earlier statement that the elements of the continued fraction being integers. I could go on about that entry, but I wont. Suffice it to say that today in mathematics, generalized continued fractions refer to several different things, among them: matrix continued fractions-- see the work of Levrie, and branched or branching continued fractions. 
I wish I had time to write a complete entry on continued fractions, but I do not. 
Here are some book suggestions for those enthusiastic about updating this entry. PLEASE have a look at the following books:
The best reference on the subject is:
1) "Continued Fractions with Applications", by Lisa Lorentzen and Haakon Waadeland, North Holland, 1992. (This covers analytic theory and a bit of arithmetic theory.)
On the arithmetic side, a great reference is: 
2) "Continued Fractions" by Andrew M. Rockett and Peter Szusz. World Scientific, 1992. (This is a relatively up-to-date treatment of arithmetic theory.) 
Older but good references are: 
3)  "Continued Fractions  Analytic Theory and Applications", by William B. Jones and W.J. Thron. Addison-Wesley, 1980. (Analytic theory and history covered here.)
4) "Analytic Theory of Continued Fractions" by H.S. Wall. Chelsea, 1973. (Analytic theory).
5) "Die Lehre Von Den Kettenbruchen" Band I, II,  by Oskar Perron, B.G. Teubner, 1954. (These two volumes cover both analytic and arithmetic theory and were written by the world's foremost expert on continued fractions of that day, also, I might add, a mathematician of great renown.)
Well, take care all. I am happy that people have taken the trouble to write an entry in wikipedia on continued fractions, a subject that has been near and dear to my heart for over 26 years.
:Hi,  yes,I agree the definition should be expanded.  And several more quick remarks:
* Please sign & date, don't post anonymously.
* If this has really been near-n-dear for 26 years, surely you have a few hours to spare for your true love? 
* I'll copy the references over.
linas 03:22, 5 Jan 2005 (UTC)
replies: text: There is such an algorithm, but you can do arithmetic on continued fractions without first turning them into fractions.  I have been meaning to write a Wikipedia article about this for some time.  Until then, you could take a look at [Arithmetic With Continued Fractions], which also has references to more detailed explanations. Dominus 16:02, 6 Apr 2005 (UTC)
thread_title:  Need for an expanded definition 

message: "Longer expressions are defined analogously."  Second sentence.  It looks to me that the scheme already suggests an infinitely long expression.
I think that the article is becoming inconsistent.   For example, Theorem 1... x represents any real, but in the lead paragraph the cascade of denominators (of which x is one) is to consist of positive integers.   And I think that "x" to this point has had a special meaning, namely, it's the number which is generating the fraction.   A symbol other than x (like "k") would perhaps be better, and I believe that having "it" be a positive integer suffices both for consistency and the generality of the result.   For consistency, the first comma should be a semicolon ... [a_0; a_1, ... ]
Theorem 4, I think, needs some work.  Again, "x" seems to represent a fixed quantity generating the continued fraction in most of the article.   To me, this meaning for "x" is necessarily what the Corollaries to Theorem 4 are referring to... if you think about an infinite continued fraction with the convergents see-sawing above and below x (this being the limit and number that has generated the fraction) and gradually converging to x (and only x), then this interpretation of x is the only one which makes the corollaries true as written.   So, the choice of the letter x to represent something else in this theorem and the corollaries is perhaps unfortunate on the one hand, and leads to falsehood on the other (unless the fraction is a finite one, terminating with the quotient a_n). 
(Thinking out loud for a minute... I'm not sure yet that Theorem 4 is true, for either interpretation of "x".   Supposing the "x" means the limiting value (= the number generating the infinite continuing fraction = the continuing fraction)... one may approach a value in a see-saw fashion without having the "too small and getting bigger" things related much to the "too big and getting smaller" things in terms of closeness to the limiting value... though some of the inequalities may have implications I don't yet see.  That is... in principle at least... even if the convergents are getting closer to one another, they may be approaching the limit in a rather lopsided fashion... with "see"-ing being far more pronounced than "saw"-ing, to continue the visual metaphor.   For the other meaning of "x", I'm not sure either.   One meaning may be true, one false.   I'll have to really think about this, of course.   As for this thinking... it seems to me that Theorem 5 has considerable bearing, and in fact Theorem 5 might be better placed before Theorem 4.)
I've added the word "nontrivial" to the parenthesized proof of Corollary 1 to Theorem 3... I don't think this is controversial.
I'm not following the text for Semiconvergents at all (but the concept is interesting).   What exactly is to be between what?   I would guess that ... the semiconvergent fraction, the one involving the a's, is to be between the successive convergents, the ones explicitly mentioned.   What exactly is being claimed ... what exactly is the theorem?
Pell's equation... no, p=1 and q=0 is a solution.   I'll need to think before editing this, for now I'll just edit to indicate both p and q are positive.
I've noted that there are actually four, rather than three, convergents displayed in the Infinite Continued Fractions section, and that these may be considered numbers 0, 1, 2, and 3.
(I've returned to this to edit, expand, and tidy up some grammar... Mathguy2, later in the day, Nov 5 2005)
Mathguy2 20:08, 5 November 2005 (UTC)
thread_title:  General comments 

message: Can someone supply a brief section on the history of continued fractions?--Niels Ø 16:24, 19 February 2006 (UTC)
thread_title:  History of continued fractions 

message: I've read through the discussion of this page, and I notice that "q-analogue" has already pointed this out. The definition of a continued fraction in this article is ridiculously restrictive. I imagine that whoever set this up in the first place wanted to avoid dealing with the very messy problem of convergence – that's a lot tougher than dealing with the convergence of series, especially when the cf is taken over the complex numbers.
Anyway, I'm intending to fix this definition fairly soon, and also in the associated article generalized continued fraction. I want to add more content about continued fractions, and I really think the definition in this article has to be fixed, or future authors won't be able to build on it. DavidCBryant 12:52, 30 November 2006 (UTC)
replies: text: AFAIK, the definition given in this article (in which the co-efficients are restricted to positive integers) is the standard definition in number theory for a simple continued fraction. Although the concept of a continued fraction can clearly be extended in various ways, we do need to be careful that when we use terms in Wikipedai we use them with their usual meanings. If you intend to extend or change the definition given in this article, please cite sources for your new definition, so we can be sure it is not original research. Gandalf61 15:47, 30 November 2006 (UTC)
replies: text: Citing a few references is not enough to justify changing the definition. I think the article should have a sort of disambiguation near the top, pointing out that "continued fraction" has several more-or-less restrictive/inclusive meanings in different mathematical fields, and pointing the reader to the article covering the desired meaning. Then, this article should go on to describe the most common meaning.--Niels Ø 18:13, 30 November 2006 (UTC)

text: To DavidCBryant: You said "...well-formed continued fractions for functions like ln and sin and arctan converge a whole lot faster than the best of their series counterparts.". I would like to know more about this. I was working on the pages for Natural logarithms and Inverse trigonometric functions and I was disappointed at how slowly the series converge. JRSpriggs 07:33, 6 December 2006 (UTC)
thread_title:  Overly restrictive definition 

message: I added an assessment of this article as "Start" class, importance "High". The reason I didn't rate it higher on the quality scale was that, although it seems to have an appropriate depth of content, the organization of that content is lacking. In addition I found the motivation section shallow; to my mind the application to diophantine approximation is what gives this subject its high importance, but one has to pore through the whole article to find that. —David Eppstein 04:44, 8 February 2007 (UTC)
thread_title:  Assessment 

message: In "Best rational approximations":
Added subscripts to clarify which term was being used in the expressions.  Otherwise it would be difficult to be sure which term of series is meant.
JKSellers 18:07, 5 March 2007 (UTC)
thread_title:  Clarification 

message: I just changed a subtitle "Golden ratio is the most irrational number" to "A property of the golden ratio φ". I had two reasons. First, it's hyperbole; and second, it's not true. Even ignoring the fact that no irrational number can be "more irrational" than another one, the heading was misleading because there are in fact infinitely many real numbers whose regular continued fraction expansions end in the string […1, 1, 1, …]. See this article.
:: Sorry, I was carried away by the hyperbole. :) Ricardo sandoval 19:07, 24 April 2007 (UTC)
replies: text: There's no need to apologize. It's actually a pretty cute way to describe the √5/(5n2) limitation. Maybe it can be rewritten somehow, so that it's not just cute, but also true.  ;^>  DavidCBryant 19:46, 24 April 2007 (UTC)
thread_title:  Lagrange spectrum 

message: By the topic of "Infinite continued fractions", I found the convergent of [a0;a1,a2,a3...]should be like this:
a0, (a0a1 + 1)/a1, [a0(a1a2 + 1) + a2] / (a1a2 + 1), {a0[a1(a2a3 + 1) + a3] + (a2a3 + 1)} / [a1(a2a3 + 1) + a3]
>because:
a0 + 1 / (a1 + 1 / a2)
=a0 + 1 / [(a1a2 + 1) / a2]
=a0 + a2 / (a1a2 + 1)
=[a0(a1a2+1) + a2] / (a1a2 + 1)
>and this:
a0 + 1 / [a1 + 1 / (a2 + 1 / a3)]
= a0 + 1 / {a1 + 1 / [(a2a3 + 1) / a3]}
= a0 + 1 / {a1 + a3 / (a2a3 + 1)}
= a0 + 1 / {[a1(a2a3 + 1) + a3] / (a2a3 + 1)}
= a0 + (a2a3 + 1) / [a1(a2a3 + 1) + a3]
= {a0[a1(a2a3 + 1) + a3] + (a2a3 + 1)} / [a1(a2a3 + 1) + a3]
Meavel 10:17, 28 May 2007 (UTC)
thread_title:  Typing error 

message: By the topic of "Calculating continued fraction representations", I think I have a better way to do this. 
Because fractions is better than decimals.
Solution 1 3.245 = 3 + 245 / 1000 = 3 + 1 / (1000/245) = 3 + 1 / (4 + (20 / 245)) = 3 + 1 / (4 + (1 / (245 / 20))) = 3 + 1 / (4 + (1 / (12 + (5 / 20)))) = 3 + 1 / (4 + (1 / (12 + (1 / 4)))) =[3;4,12,4]
OR
Solution 2
by calculation of 3.245...
1000/245 = 4 + 20/245; 245/20 = 12 + 5/20; 20/5 = 4
3.245 = [3;4,12,4]
Better than the original page: 3.245-3=0.245, 1/0.245=4.082,4.082-4=0.482,1/0.482...
Meavel 10:20, 28 May 2007 (UTC)
thread_title:  Suggestion 

message: By the topic of "Infinite continued fractions", I found the convergent of [a0;a1,a2,a3...]should be like this:
a0, (a0a1 + 1)/a1, [a0(a1a2 + 1) + a2] / (a1a2 + 1), {a0[a1(a2a3 + 1) + a3] + (a2a3 + 1)} / [a1(a2a3 + 1) + a3]
>because:
a0 + 1 / (a1 + 1 / a2) =a0 + 1 / [(a1a2 + 1) / a2] =a0 + a2 / (a1a2 + 1) =[a0(a1a2+1) + a2] / (a1a2 + 1)
>and this:
a0 + 1 / [a1 + 1 / (a2 + 1 / a3)] = a0 + 1 / {a1 + 1 / [(a2a3 + 1) / a3]} = a0 + 1 / {a1 + a3 / (a2a3 + 1)} = a0 + 1 / {[a1(a2a3 + 1) + a3] / (a2a3 + 1)} = a0 + (a2a3 + 1) / [a1(a2a3 + 1) + a3] = {a0[a1(a2a3 + 1) + a3] + (a2a3 + 1)} / [a1(a2a3 + 1) + a3]
Somebody who knows TEX language please correct it. I talked it for two months.
--Meavel 06:48, 31 July 2007 (UTC)
replies: text: I'm sorry, Meavel, but your formula is wrong. The formula in the article is correct. Please consult the references if you want to see a proof. Or write to me on my talk page and I'll show you the correct algebra (proof by induction) over there.  DavidCBryant 12:09, 31 July 2007 (UTC)
thread_title:  attention 

message: Clearly a section about the comparison of two continued fractions is missing.
If that section would be added the section about best approximations could be shortened and made more clear.
Hkleinnl (talk) 11:46, 18 December 2007 (UTC)
thread_title:  missing entry: comparison of two continued fractions. 

message: At the bottom of the section titled "Calculating continued fraction representations" the following appears:
:The number 3.245 can also be represented by the continued fraction expansion [3; 4, 12, 3, 1]; refer to Finite continued fractions below.
:This algorithm is suitable for real numbers, but can lead to numerical disaster if implemented :with floating point numbers. Instead, any floating point number is an exact rational (the :denominator is usually a power of two on modern computers, and a power of ten on electronic :calculators), so a variant of Euclid's GCD algorithm can be used to give exact results.
As currently phrased, this is confusing. I'm revising it to more accurately reflect the problem and the correct solution. Solemnavalanche (talk) 18:46, 23 December 2007 (UTC)
thread_title:  floating point numbers 

message: "One theorem states that any real number k can be approximated by rational m/n " .. I'd like to see a reference for this, and the name of the theorem cited.  —Preceding unsigned comment added by 99.226.209.46 (talk) 05:19, 6 January 2008 (UTC)
replies: text: It's Theorem 193 in Hardy & Wright (5th ed.) - I've added a reference in the article. Gandalf61 (talk) 17:23, 6 January 2008 (UTC)
thread_title:  citation needed 

message: As an additional point of interest, the fact that all rational numbers can be represented as a finite continued fraction allows mathematical induction to be extended to the positive rationals. This can be done by proving:
1. n=1 is true
2. If n=k is true, then n=k+1 is true
3. If n=k is true, then n=1/k is true
Using these operations, it is possible to create all finite continued fractions and hence all rational numbers. Perhaps this deserves a mention on either this page or the page on induction?
--124.189.100.175 (talk) 03:03, 12 January 2008 (UTC)
replies: text: That can be done without continued fractions. E.g., consider a grid of points in a coordinate system, e.g. {..., -3, -2, -1, 0, 1, 2, 3, ...} X {1, 2, 3, ...}. Define a sequence going through all these points, beginning at (0, 1), spiralling or zigzagging outwards as you please. Let each point (p, q) represent the fraction p/q, and skip it if it is not to lowest terms. Of course, this skipping in some contexts may makes this sequence inferior to the one suggested by 124.189.100.175, so I guess what I'm saying is, if we include 124.189.100.175's sequence in the article, we should indicate what it might be good for.--Niels Ø (noe) (talk) 11:23, 12 January 2008 (UTC)
thread_title:  Rational induction 

message: Every finite continued fraction represents a rational number, and every rational number (except 1) can be represented in precisely two different ways as a finite continued fraction.
 
Why except 1? 1 can also be represented in two different ways : 1 = [1] = [0;1]. Indeed any integer n = [n] = [n-1;1].--143.248.181.36 (talk) 06:48, 28 January 2008 (UTC)
thread_title:  representation of 1 

message: Is Theorem 2 really a Theorem?  All it is, is the definition of h and k, which is a given. TechnoFaye Kane 03:28, 8 February 2008 (UTC)
replies: text: As the article stands, hn and kn are defined at the beginning of the Some useful theorems section as integer sequences with recurrence relations that use the an as parameters. So, yes, it is then necessary to prove that hn/kn is the nth convergent of the continued fraction. An alternative approach is to reverse the order, defining hn and kn as the numerator and denominator of the nth convergent, and then showing that hn and kn satisfy the recurrence relations. Gandalf61 (talk) 11:17, 8 February 2008 (UTC)
replies: text: THANK YOU!!!! TechnoFaye Kane 20:46, 8 February 2008 (UTC)
thread_title:  Theorem 2 

message: I saw this at the beginning of the first section, "Motivation", and it made me laugh:
:Most people are familiar with the decimal representation of real numbers:
::r = \sum_{i=0}^\infty a_i 10^{-i}
It looks as if it's being said that most people understand the above mathematical representation. Freaked me out.
(This entry edited once for my own reasons. New sig below)
- Misha Vargas
216.254.12.114 (talk) 04:08, 21 November 2008 (UTC)
replies: text: This is only true. Most people are familiar with decimal representation, even if not in the rigorous form. Scineram (talk) 11:38, 17 December 2008 (UTC)
thread_title:  Nerd Bias 

message: The opening section introduces the terminology of partial numerators and partial denominators that pertain to generalized continued fractions.  But we fail to mention that for the simple or regular continued fractions dealt with in this article (having all partial numerators 1) the positive integers ai are termed partial quotients.  Cf. the linked Wikipedia article on "restricted partial quotients".
My suggestion is to introduce the partial quotients terminology in the opening and to make the topic of generalized continued fractions a subsection (as is often done with generalizations of a math concept).  Hardmath (talk) 12:15, 27 May 2009 (UTC)
thread_title:  Terminology of partial quotients 

message: Let's discuss about the formula of this picture:
:
\frac{a_0}{1},\qquad
\frac{a_1a_0+1}{a_1},\qquad
\frac{    a_2(a_1a_0+1)+a_0}{a_2a_1+1},\qquad
\frac{a_3(a_2(a_1a_0+1)+a_0)+(a_1a_0+1)}{a_3(a_2a_1+1)+a_1}.
from infinite continued fraction of continued fraction. Continued_fraction
please compare with this.
:x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{\ddots\,}}}} 
and here is my proof to show that conflicts.
Answer:Covergents:
:x_1 = [a_0] = a_0\qquad
:x_2 = [a_0;a_1] = a_0 + \cfrac{1}{a_1} = \frac{a_1a_0+1}{a_1}\qquad
:x_3 = [a_0;a_1,a_2] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2}} = a_0 + \cfrac{1}{ \cfrac{a_1a_2 + 1}{a_2}} = a_0 + \cfrac{a_2}{a_1a_2 + 1} = \cfrac{a_0(a_1a_2 + 1) + a_2}{a_1a_2+1}\qquad
:x_4 = [a_0;a_1,a_2,a_3] = a_0 + \cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3}}}\qquad
:= a_0 + \cfrac{1}{a_1+\cfrac{a_3}{a_2a_3 + 1}}\qquad
:= a_0 + \cfrac{a_2a_3 + 1}{a1(a_2a_3 + 1) + a_3}\qquad
:= \cfrac{a_0(a_1(a_2a_3 + 1) + a_3) + a_2a_3 + 1}{a_1(a_2a_3 + 1) + a_3}\qquad--Meavel (talk) 00:26, 17 March 2008 (UTC)
replies: text:  As far as I can see the formulae are only seemingly different, but in fact the same Hkleinnl  —Preceding comment was added at 20:00, 19 March 2008 (UTC)
replies: text:  Yes, if you expand the numerator and denominator in each case, you will find that the two match. Asmeurer (talk  ♬ contribs) 17:27, 13 September 2009 (UTC)
thread_title:  infinite continued fraction 

message: The formula given on the page does not produce the correct continued fraction. Using this formula for k=0, 1, ... and n=5 yields
[1; 2, 30, 12, 1, 1, 5 ...]
But Wolfram suggests http://www01.wolframalpha.com/input/?i=exp+%282%2F5%29 
[1; 2, 30, 12, 1, 1, 17, 90, 27, 1, 1, 32, 150, 42, 1, 1, 47, 210, 57, 1, ...] 
Wolfram's CF also collapses to an approximation that agrees with all floating point approximation formulas (scheme, google, etc). I cannot find a reference that represents this sequence as a simple continued fraction elsewhere. Does anyone have the correct formula and a reference?
Erislover (talk) 04:00, 18 July 2009 (UTC)
replies: text: The formula should have read "3nk" instead of "3k" and has now been corrected. Thanks. Glenn L (talk) 04:31, 18 July 2009 (UTC)
thread_title:  exp(2/odd) incorrect 

message: Could anybody give an exact reference of that fact, please? I cannot find it in any book. Thanks. 147.83.133.93 (talk) 09:17, 28 July 2009 (UTC)
replies: text: Have you looked in Khinchin?  I would be very surprised if it were not there.  —Dominus (talk) 14:43, 28 July 2009 (UTC)

text: It does follow immediately from theorem 4 on page 6 (for finite continued fractions) and theorem 8 on page 9 (for infinite continued fractions). —Dominus (talk) 14:46, 28 July 2009 (UTC)
replies: text: Are you sure? Theorem 4 on page 6 is equivalent to Theorem 4. But I'm not sure that comparison of continued fractions' result is an immediat consequence of that. Any ideas? Thank you. 147.83.133.93 (talk) 10:46, 29 July 2009 (UTC)
thread_title:  Comparison of continued fractions 

message: This one of 5 useful theorems stated without proof. Why not state what is more basic in this case. Something like this:
Theorem 4
The even convergents continually increase and the odd convergents continually decrease. Also, every even convergent is less than every odd convergent. In symbols
a_0<a_2<a_4<a_6<...<a_7<a_5<a_3<a_1
Furthermore, each convergent is closer to the next than the previous to it was . In other words, in the decreasing positive sequence a_1-a_0 > a_1-a_2 > a_3-a_2 > a_3-a_4 > a_5-a_4 > ... each term is less than half the previous one. 
Combining these:
a_0< \frac{a_0+a_1}{2}<a_2< \frac{a_2+a_3}{2}<a_4< \frac{a_4+a_5}{2}<a_6<...<a_7< \frac{a_5+a_6}{2}<a_5< \frac{a_3+a_4}{2}<a_3< \frac{a_1+a_2}{2}<a_1
Corollary 1: For an infinite (simple positive) continued fraction, there is exactly one real number a which is greater than all the even convergents and less than all the odd ones.
--Gentlemath (talk) 07:14, 4 October 2009 (UTC)
thread_title:  suggestion for theorem 4 

message: What is meant by?:
:One formal description of the half rule is that the halved term, 1⁄2 ak, is admissible if and only if
::\left[a_k;a_{k-1},\ldots,a_1\right] > \left[a_k;a_{k+1},\ldots\right].
Is it really correct that the indexes are starting with a_k and going down with the first elipses?  In all other cases the indexes increase as we go to the right.  Also, are the indexes going up with the second elipses, with the implication that we take the limit as the number of terms goes to infinity?  Thanks! Quantling (talk) 17:03, 31 March 2010 (UTC)
thread_title:  the "half rule" 

message: What is the rational with the smallest denominator that falls within the interval (x,y), [x,y), (x,y], or [x,y]?  If both x and y are irrational the four cases are identical; and with
:\begin{align}
x &= [a_0; a_1, a_2, \ldots, a_{k-1}, a_k, a_{k+1}, \ldots ]\\
y &= [a_0; a_1, a_2, \ldots, a_{k-1}, b_k, b_{k+1}, \ldots ]
\end{align}
where x and y have identical continued fraction expansions up through a_{k-1}, I believe the answer is equal to 
:z = [a_0; a_1, a_2, \ldots, a_{k-1}, \min(a_k,b_k)+1]\,.
If x is rational, one has to take care to choose the appropriate one of the two representations for x; in some sense one uses the continued fraction expansion of either x+\epsilon or x-\epsilon, depending upon whether the lower bound is more like an irrational of the form x+\epsilon or an irrational of the form x-\epsilon.  Specifically, for a rational within (x,y) or within (x,y], the lower bound is, in some sense, x+\epsilon; one should use [a_0; a_1, ..., a_{k-1}, a_k] if k is even; one should use [a_0; a_1, ..., a_{k-1}, a_k - 1, 1] if k is odd.  If one wants a rational within [x,y) or [x,y] then the lower bound is more like x-\epsilon, and the choice for the appropriate expansion of x is the other way around.
Similarly, if y is rational, for a rational within (x,y) or within [x,y), the upper bound is, in some sense, y-\epsilon; one should use [b_0; b_1, ..., b_{k-1}, b_k] if k is odd; one should use [b_0; b_1, ..., b_{k-1}, b_k - 1, 1] if k is even.  If one wants a rational within (x,y] or [x,y] then the upper bound is more like y+\epsilon, and the choice for the appropriate expansion of y is the other way around.
With these choices one uses the formula above applicable to irrational numbers, unless these choices leave the expansions for x and y equal up to the length of the shorter one.  In this last case, truncate the longer one to one term more than the shorter one, incrementing the very last value in the truncated sequence.
If this is truth does anyone have a citable source? Quantling (talk) 19:27, 12 April 2010 (UTC)
replies: text: You are right, at least in the case of irrationals (for rational boundaries I didn't check; it is somewhat messy in any case). To understand such questions, look at Stern–Brocot tree, to which I just added a link and corrected the contents for you convenience. In fact a better way to think of a path in the Stern–Brocot tree is a a sequence of left/right decisions rather than continued fractions; a coefficient ai corresponds to that many successive decisions in the same direction (after which the direction changes), except for the final coefficient which gives one less such decision. Then to compare two paths find the largest common prefix; if both paths continue beyond that point (in different directions) then this gives the simplest fraction in the interval. If only one path continues, then either take the common prefix if the interval was closed at this end, if not continue one step in the direction of the longer path if that gives a valid point of the interval, and if not (an open interval of which we just visited both boundaries) continue one step in the opposite direction of the last step. It is straightforward (though a bit cumbersome) to translate all this back into the language of continued fractions. Marc van Leeuwen (talk) 10:09, 13 April 2010 (UTC)
thread_title:  Best rational within an interval 

message: Would another benefit over decimal representations worth mentioning be that some numbers do not have unique decimal representations (when the decimal ends in an infinite string of 9's)?  Asmeurer (talk  ♬ contribs) 17:23, 13 September 2009 (UTC)
replies: text: Isn't it the case that every rational does not have a unique continued fraction representation?  That is, [a_0; a_1, \ldots, a_k, 1] always equals [a_0; a_1, \ldots, a_k+1], yes?  In this sense continued fraction representations are worse than decimal representations; for the latter not all rational numbers have non-unique representations. Quantling (talk) 18:04, 31 March 2010 (UTC)
thread_title:  Non-unique decimal representation 

message: I wonder what makes a person so "priviliged" that his/her external-link is allowed to  appear at wikipedia. I mean, if Wikipedia does not like EXTERNAL LINKS then WIKIPEDIA SHOULD NOT mantain the external-links section. I think it is an obvious public-outrage for many people to realize that only two or three guys are "priviliged" by wikipedia, i do not see any justice on that. Wikipedia should either mantain the EXTERNAL LINKS in another related page or definitely eliminate all of them. 
What makes some few guys so "privileged"?
What makes some guy to be the judge for selecting only two or three external links?
Mirificium (talk) 22:47, 20 March 2010 (UTC)
thread_title:  EXTERNAL LINKS 

message: I'm often astonished about how people can be at ease with definitions that don't even begin to make sense. The current lede is an example: "In mathematics, a continued fraction is an expression such as
:x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \ddots\,}}}} "
With the dots in there, it is not an expression because part of it is not specified (how should one evaluate such an expression)? If the dots stand for infinite continuation in a similar fashion (as the term "continued" suggests), it is not an expression either, because (1) it is infinite (not much hope for evaluating such an expression), and (2) viewing it as an infinitely nested tree structure, there is no place to even begin evaluation of the expression. So no, a continued fraction is not an expression of that kind, the continued faction associated to a sequence of integers is a sequence of rational numbers, defined using expressions similar to the one depicted, but without dots (or by recurrence). Also I see no clear mention in the article that all infinite continued fractions actually designate a well defined real number, in other words a convergence statement. I'll make a stab at giving a more sensible definition and description, while remaining coherent with what is said later in the article. Marc van Leeuwen (talk) 08:50, 2 April 2010 (UTC)
replies: text: Good point!  Since you didn't edit the article right away, I took a shot.  If my edits collide with yours, feel free to overwrite mine, or merge, or whatever. Quantling (talk) 14:04, 2 April 2010 (UTC)
thread_title:  Improper definition 

message: How do these changes to the table look?
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Calculating continued fraction representations
Consider a real number r.
Let i be the integer part and f the fractional part of r.
Then the continued fraction representation of r is [i; a1, a2,...], where [a1; a2,...] is the continued fraction representation of 1/f. 
To calculate a continued fraction representation of a number r, write down the integer part (technically the floor) of r.  Subtract this integer part from r.  If the difference is 0, stop; otherwise  find the reciprocal of the difference and repeat.  The procedure will halt if and only if r is rational.
Find the continued fraction for 3.245                  	Find the continued fraction for 3.245                  	Find the continued fraction for 3.245                  	Find the continued fraction for 3.245                  	Find the continued fraction for 3.245                  	Find the continued fraction for 3.245
Real Number                                            	Integer portion (floor)                                	Fractional portion                                     	Simplified fraction                                    	Reciprocal of f                                        	Simplification of 1/f
r = 3.245\,                                            	i = 3\,                                                	f = 3.245\ \left(3 \tfrac{49}{200}\right) - 3\,        	= 0.245\ \left(\tfrac{49}{200}\right)\,                	1/f = 1 / 0.245\ \left(\tfrac{200}{49}\right)\,        	= 4.082\ \left(4 \tfrac{4}{49}\right)\,
r = 4.082\,                                            	i = 4\,                                                	f = 4.082\ \left(4 \tfrac{4}{49}\right)- 4\,           	= 0.082\ \left(\tfrac{4}{49}\right)\,                  	1/f = 1 / 0.082\ \left(\tfrac{49}{4}\right)\,          	= 12.250\ \left(12 \tfrac{1}{4}\right)\,
r = 12.250\,                                           	i = 12\,                                               	f = 12.250\ \left(12 \tfrac{1}{4}\right)- 12\,         	= 0.250\ \left(\tfrac{1}{4}\right)\,                   	1/f = 1 / 0.250\ \left(\tfrac{4}{1}\right)\,           	= 4.000\ \left(4\right)\,
r = 4.000\,                                            	i = 4\,                                                	f = 4.000\ \left(4\right) - 4\,                        	= 0.000\,                                              	STOP                                                   	
Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]
3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}
The number 3.245 can also be represented by the continued fraction expansion [3; 4, 12, 3, 1]; refer to Finite continued fractions below.
This algorithm is suitable for real numbers, but the limited precision of floating point numbers will often lead to small errors, skewing the final result. Instead, floating point numbers should be converted to rational numbers.  The denominator is usually a power of two on modern computers, and a power of ten on electronic calculators, so a variant of Euclid's GCD algorithm can be used to give exact results.
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I think that this will be much easier to read and understand. The original version was like this:
Find the continued fraction for 3.245                  	Find the continued fraction for 3.245                  	Find the continued fraction for 3.245                  	Find the continued fraction for 3.245                  	Find the continued fraction for 3.245
3\,                                                    	3.245\ \left(3 \tfrac{49}{200}\right) - 3\,            	= 0.245\ \left(\tfrac{49}{200}\right)\,                	1 / 0.245\ \left(\tfrac{200}{49}\right)\,              	= 4.082\ \left(4 \tfrac{4}{49}\right)\,
4\,                                                    	4.082\ \left(4 \tfrac{4}{49}\right)- 4\,               	= 0.082\ \left(\tfrac{4}{49}\right)\,                  	1 / 0.082\ \left(\tfrac{49}{4}\right)\,                	= 12.250\ \left(12 \tfrac{1}{4}\right)\,
12\,                                                   	12.250\ \left(12 \tfrac{1}{4}\right)- 12\,             	= 0.250\ \left(\tfrac{1}{4}\right)\,                   	1 / 0.250\ \left(\tfrac{4}{1}\right)\,                 	= 4.000\ \left(4\right)\,
4\,                                                    	4.000\ \left(4\right) - 4\,                            	= 0.000\,                                              	STOP                                                   	
Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]     	Continued fraction form for 3.245 is [3; 4, 12, 4]
3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}	3.245 = 3 + \cfrac{1}{4 + \cfrac{1}{12 + \cfrac{1}{4}}}
By the way, previous edits by 65.190.71.252 were by me as well. Pisharov (talk) 20:07, 2 June 2010 (UTC)
thread_title:  Table @ "Calculating continued fraction representations" 

message: Would someone please be so friendly and explain where in this Krylov space method the continued fractions occur? -- LutzL (talk) 10:03, 29 July 2010 (UTC)
thread_title:  Lanczos algorithm -- Eigenvalues and eigenvectors 

message: i'm not sure wether this is appropriate here, so feel free to remove the link which I've just added.
In the context of fractional iteration( and tetration) I consider the fraction 1/(1+x) as iterable, so as finite and infinite continued fraction. Maybe this link to fractional iteration has some informative value in the article here (generalizations )
Feel free to remove the link if not appropriate -
Gottfried Helms--Gotti 15:23, 1 August 2010 (UTC)
thread_title:  fractional iteration of continued fraction