# Talk:Generalized continued fraction

## Incorrect formatting in series section

The LaTeX in the series section is all messed up, starting with Euler's formula. I don't know this stuff well so I would prefer to leave it to somebody else to fix. I think frac needs to be changed to cfrac, and parentheses need to be moved around. Adking80 (talk) 14:29, 29 June 2009 (UTC)

## Bug

Would somebody familiar with this topic comment on the Bug here thing inserted by somebody in the main text? Thanks a lot. Oleg Alexandrov 05:02, 11 Apr 2005 (UTC)

Didn't have time to read it, but two things I noticed right away:
1. the usual terminology is "simple continued fraction" (not "c.f." unless qualified by "hereafter we deal only with s.c.f.s") and "continued fraction" (not "generalized c.f.")
2. Many, many "generalized continued fractions" exist, but this usually refers to something like the Jacobi-Perron algorithm for simultaneous rational approximation of real vectors, or for continued fractions in which the coefficients are functions, for operator-valued formulations, etc.
A good place to begin reading about one dimensional continued fractions is
• {{#invoke:citation/CS1|citation

|CitationClass=book }}, but this book is already seriously out of date.

HTH ---CH 21:10, 5 May 2006 (UTC)

Well, I'm sorry I won't get a chance to talk to Mr. Hillman. It appears that he's not participating in Wikipedia any longer. But what he said goes double for me. Anyway, the "Bug here" thing that somebody once inserted undoubtedly referred to the absolutely ludicrous definitions currently in place for "continued fraction" (in this article) and "generalized continued fraction" (in this one). DavidCBryant 20:21, 30 November 2006 (UTC)

## Not "generalized" enough?

It is absolutely incorrect to use the name "Generalized Continued Fraction" for second-order continued fractions as those defined here. All this is explained at: Generalized Continued Fractions —Preceding unsigned comment added by Arithmonic (talkcontribs) 02:43, 15 October 2006

Wikipedia is not about what's correct and what's incorrect. It's about what the term "Generalized Continued Fraction" refers to in the mainstream mathematical literature. -- Jitse Niesen (talk) 04:12, 15 October 2006 (UTC)
Well, I've read quite a few books about continued fractions, and I have never seen a reference to a "generalized" continued fraction whose partial numerators and partial denominators have to be integers. I'm absolutely certain that arithmonic is right. The thing defined by this article right now is a very special case of a very narrow class of continued fractions. Saying that it's "generalized" is farcical. DavidCBryant 20:21, 30 November 2006 (UTC)
Are you saying that the references listed in the article do not define "generalized continued fraction" as explained in the article? Then, I'd be very grateful if you could change it. If, instead, you're saying that the list of references is not representative, then add your definition and quote a reference which uses it. By the way, the TeX command for ${\displaystyle \ddots }$ is \ddots (diagonal dots). -- Jitse Niesen (talk) 01:50, 1 December 2006 (UTC)
Yeah, I was saying that the definition that was in the article is not generally accepted by mathematicians, including the mathematicians who wrote the references. So I've made some initial changes, JN. I've also added one new article about the fundamental recurrence relationship between successive convergents, and I've added some content to the article about the convergents of a continued fraction.
So far I'm just trying to open up the number-theoretic definition that used to be here so that applications of continued fractions to complex analysis are possible, by the "Wikipedia" definition. Other generalizations (into continued fractions of differential operators, or continued fractions of matrices) are a little outside my area of expertise, but I'll try to work in a couple of references to them. DavidCBryant 16:14, 3 December 2006 (UTC)

## Inconsistent symbols for partial denominators

I notice that the article Continued fraction uses "an" for partial denominators while this article Generalized continued fraction uses that for the partial numerators and uses "bn" for partial denominators. I feel that these should be changed to be consistent. JRSpriggs 06:20, 15 December 2006 (UTC)

I'd really rather not do that. The usage here (a in the numerator and b in the denominator, with the somewhat weird 0th denominator standing outside the fraction) is practically universal in the analytic theory of continued fractions, from Gauss until the present day. If we do what you're proposing, it will probably just confuse anyone who is actually trying to study analysis and continued fractions out of a book.
Continued fractions have applications in both number theory and analysis (maybe in algebra, too -- I'm not sure about that). The number theory guys don't care about fractions that incorporate anything except integers. In analysis we can get all kinds of weird things floating around inside the fraction. The notation has evolved differently pretty naturally, because in number theory the partial numerators are almost always "1", and can effectively be ignored, while in analysis we prove a theorem (not in Wikipedia yet, but I'm working on it) that "most" continued fractions can be expressed in infinitely many ways, two of the most useful of which involve "1"s as numerators or, alternatively, "1"s as denominators. In analysis we can even have some 0s in the denominators and still get good (convergent) results.
Anyway, the "continued fraction" article is already dedicated to number theory (and not even the more interesting aspects of that, like solving Diophantine equations; it's mostly about the representation of irrational numbers in canonical form). Can't this article be for analysis? DavidCBryant 12:59, 15 December 2006 (UTC)
You could change the an in the other article to bn instead. I did not say anything about number theory versus analysis. JRSpriggs 08:43, 16 December 2006 (UTC)
I read too much into your remarks, I suppose (arithmetic vs analysis). The thing is, I have run across both notations, in books. And it is fairly common for books that concentrate on the number theoretic applications of continued fractions (such as Diophantine equations, and how the set of all Dedekind cuts can be placed in one-to-one correspondence with the simple cfs in canonical form) to use the symbol a in the partial denominators; and for books that concentrate on analytic cfs to use the a/b convention that's in this article right now. Anyway, when I first suggested trying to introduce a more general definition into the continued fractions article, I felt as if I had a tiger by the tail! I'm not sure I want to re-open that can of worms.
I'm not opposed to making changes to articles. But this inconsistency in notation is present in the mainstream literature, AFAIK. So I don't see much harm in leaving a similar inconsistency in Wikipedia. But yeah, if one were determined to eliminate the inconsistency, I'd rather do it by opening up the continued fractions article just a little bit. DavidCBryant 12:02, 16 December 2006 (UTC)

## History section

I'm particularly curious about one thing. I'm fairly certain that the uniqueness of the canonical continued fraction representation of an irrational real number was not a subject of discussion until sometime during the 19th century. It just makes sense that this result would have been roughly contemporaneous with Kronecker's research into Cauchy sequences and Dedekind's new idea of a cut in the rationals. But I'm not sure who came up with the canonical representation in the first place. Lagrange used canonical fractions for quadratic irrationals, but he didn't know about transcendental numbers, which didn't really become a hot issue until algebraic number theory was fairly far along. Does anybody have a clue, or an idea of how I should search for this factoid? DavidCBryant 01:06, 16 December 2006 (UTC)

## Regularization of generalized continued fractions

If one multiplies both the numerator and denominator of one of the ratios by the same constant c, then the resulting continued fraction should have the same value. For example, if we do this to the second ratio, this

${\displaystyle x=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{\ddots \,}}}}}}}}}$

would become this

${\displaystyle x=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}c}{b_{2}c+{\cfrac {a_{3}c}{b_{3}+{\cfrac {a_{4}}{\ddots \,}}}}}}}}}$

If we then divide each ratio's numerator and denominator by its numerator, we get

${\displaystyle x=b_{0}+{\cfrac {1}{{\frac {b_{1}}{a_{1}}}+{\cfrac {1}{{\frac {b_{2}a_{1}}{a_{2}}}+{\cfrac {1}{{\frac {b_{3}a_{2}}{a_{1}a_{3}}}+{\cfrac {1}{\ddots \,}}}}}}}}}$

Should we not put this into the article? JRSpriggs 09:06, 16 December 2006 (UTC)

Yes, this should definitely go in the article. It's called an equivalence transformation, or sometimes an equivalence relation, and it's usually expressed this way:
${\displaystyle b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{\ddots \,}}}}}}}}=b_{0}+{\cfrac {c_{1}a_{1}}{c_{1}b_{1}+{\cfrac {c_{1}c_{2}a_{2}}{c_{2}b_{2}+{\cfrac {c_{2}c_{3}a_{3}}{c_{3}b_{3}+{\cfrac {c_{3}c_{4}a_{4}}{\ddots \,}}}}}}}}}$
where the ci are any non-zero complex numbers, and equality is to be understood in the sense that the two sequences of successive convergents are numerically equal, term by term. The advantage of expressing it this way is that it then becomes easy to explain the transformation into all "1"s in the numerators, as well as the transformation into all "1"s in the denominators, and some other special forms that prove useful as the theory is developed farther.
Anyway, I was aiming to put the equivalence transformation in an "elementary results" section, along with some observations on vanishing ai and bi, and maybe some stuff about separating a continued fraction into its even and odd parts. I'd welcome your assistance if you want to collaborate, JR. dcb
PS Oh, yeah, I almost forgot to mention another thing that ought to be in this article. A particular continued fraction can be thought of as the composition of a sequence (possibly infinite) of Möbius transformations. I think that idea should come right after the "elementary considerations" section, or maybe even be a subheading under "elementary considerations". The advantage of using this device is that it brings the machinery of complex analysis (conformal mapping) directly to bear on the continued fraction itself, and this makes many proofs much simpler than they would be without it. (I guess one disadvantage might be that a link to mobius transformation brings the reader directly to a "bijective conformal map" on the Riemann sphere, and to "automorphism group", as well, which will probably scare all the non-mathematicians out of town before sundown!) DavidCBryant 12:50, 16 December 2006 (UTC)

## Another formula for e

When working on the derangements at Rencontres numbers, I noticed that they yield another formula for e:

${\displaystyle e=2+{\cfrac {2}{2+{\cfrac {3}{3+{\cfrac {4}{4+{\cfrac {5}{\ddots \,}}}}}}}}}$

where the corresponding partial numerators and partial denominators are all equal and run thru the integers beginning with 2. Amazing. JRSpriggs 04:19, 21 December 2006 (UTC)

Yes, it is amazing ... and very beautiful, also. Oh -- when I looked at derangement just now I noticed a minor case of vandalism. Somebody might want to check to be sure I fixed it correctly. I'm also curious, JR -- do you think one of us ought to work a reference to continued fractions into the derangement article? DavidCBryant 12:16, 21 December 2006 (UTC)

If we can figure out how to make it sufficiently relevant to that article. JRSpriggs 08:01, 22 December 2006 (UTC)
I'll think about that a little. Pointing out that the fundamental recurrence formulas used in the theory of continued fractions are identical with the recursion relation generating the "subfactorials" might be a useful note in derangement. (I kind of like "derangements" ... mathemeticians, who in general are slightly deranged anyway, tend to like them. ;^>)
Oh -- I took a look, and there's an equivalent version of this cf in the article about Euler's number. It wouldn't hurt the section on equivalence transformations in this article to have a worked example, and this particular e-quivalence is a very natural example. Just a thought. DavidCBryant 12:07, 22 December 2006 (UTC)
I think I will put it into e (mathematical constant). And yes, showing the equivalence would be a good idea. JRSpriggs 05:29, 23 December 2006 (UTC)

## Maps whole plane to a point?

To David: You say "And every member of that automorphism group maps the extended complex plane into itself – not one of the Τns can possibly map the plane into a single point. Yet in the limit the sequence {Τn} defines an infinite continued fraction which (if it converges) does in fact map the entire complex plane into a single point.". This is impossible. Suppose w is the value on which you are trying to focus the plane. For each n, choose a point zn which maps to a point outside the disc with radius one around w. Since the extended complex plane (with the point at infinity added) is compact, this sequence must have at least one cluster point. Call it Z. Then Tn cannot converge uniformly to w in a neighborhood of Z. Is that not so? JRSpriggs 09:58, 12 January 2007 (UTC)

You're right. That is so. I have to admit I'm speaking loosely in this section of the article. I tried to emphasize this by talking about the difference between intuition and proof in the opening paragraph for the section. Talking about limits as if they can ever be "reached" is always loose talk, when you come right down to it. The point I'm trying to make is that a sequence of LFT's can have a limit (a continued fraction – not its value, but the fraction itself) that lies outside the set of all LFT's. (I guess another way to look at it is that the infinite continued fraction is not a linear fractional transformation, but each truncation of the fraction is easily associated with an LFT, and those LFT's can be thought of as forming a sequence that converges to the continued fraction.)
The analogous case that's most familiar is a Cauchy sequence of rational numbers whose limit is irrational. A simple continued fraction (with nothing but integers for elements) can always be constructed that's equivalent to such a Cauchy sequence, and vice versa. Maybe the wording should be adjusted to run more along those lines? Or maybe there should be an "intuitively" in the sentence in question, somewhere? "Does in fact" sounds pretty dogmatic.
On the other hand, I think you're talking about uniform convergence, and I was only trying to talk about simple convergence. Isn't your argument about the cluster point the same basic reason we can't really say "the power series for ez converges uniformly everywhere in the complex plane" and instead must say "this power series converges uniformly on every bounded domain lying in ℂ"? Anyway, thanks for looking at this stuff. I appreciate your feedback. DavidCBryant 12:14, 12 January 2007 (UTC)
I thought of a more concrete point, a counter-example. Consider the usual continued fraction for the golden ratio which has all the partial numerators and partial denominators equal to one. Indeed it will converge to the golden ratio as n goes to infinity for all complex z EXCEPT for ${\displaystyle {\frac {1-{\sqrt {5}}}{2}}}$ (the other root of the equation z2 - z - 1 = 0) which goes to itself regardless of what n is. JRSpriggs 12:29, 12 January 2007 (UTC)
Hi, JR! I had just about concluded that I really need to reword this thing, anyway. It's wrong to call a continued fraction a "mapping". A cf can be regarded as the limit of a sequence of "mappings", but the cf itself is not a mapping ... just as an irrational number can be described as the limit of a sequence of rationals, even though it isn't a rational number. Thanks for the very cute counter-example. Maybe we ought to work that into the article somehow? I guess it will work for any convergent periodic continued fraction with two distinct fixed points (that is, the cf will converge to one value, but the LFT actually fixes two points, one of which is the other root of the quadratic). DavidCBryant 18:34, 12 January 2007 (UTC)
Actually I made mistake in my counter-example. I should have said -- if
${\displaystyle \phi =1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{\ddots \,}}}}}}}}}$
then we get
${\displaystyle \lim _{n\to \infty }T_{n}(-\phi )=1-\phi \!}$
while
${\displaystyle \lim _{n\to \infty }T_{n}(z)=\phi \!}$
for any other extended complex number z. My error was that I was thinking of z as the entire denominator of the last ratio, but actually that is 1+z. I do not know what, if anything, we could say about such a 'point of divergence' that would be completely general. JRSpriggs 05:56, 13 January 2007 (UTC)

## set of GCF numbers

It seems that the set of numbers which can be represented by g.c.f's of which the coefficients form in a natural way a rational number (a repetitive sequence of digits) is a strict superset of the rational numbers, but of course not all of the real line. Is there a name for this set? --MarSch 18:41, 13 February 2007 (UTC)

I think I understand what you're driving at, MarSch, but I'm not certain. Are you talking about periodic continued fractions? (See this article for the definition.) If the coefficients of the continued fraction are restricted to be integers, then the superset to which you refer is the set of quadratic irrational numbers (plus the rationals, of course), or algebraic numbers of the second degree. If that restriction is removed, and the coefficients are allowed to assume arbitrary complex values, then a continued fraction can represent any complex number whatsoever.
One of the beautiful properties of continued fractions is that the (positive) square root of any non-square natural number can be represented by a periodic simple continued fraction in canonical form. So in a way, square roots are like rational numbers, because they can be represented by a finite pattern of natural numbers that repeats itself over and over again. See this article for some examples. DavidCBryant 19:24, 13 February 2007 (UTC)

## Conditions for irrationality

Can someone state the necessary and sufficient conditions for a convergent generalized continued fraction to converge to an irrational number? For instance if

${\displaystyle x=b_{0}+{\underset {i=1}{\overset {\infty }{K}}}{\frac {a_{i}}{b_{i}}}\,}$

is known to converge and

${\displaystyle a_{i}\,}$ and ${\displaystyle b_{i}\,}$ are both functions of ${\displaystyle i\,}$

and

${\displaystyle a_{i}\,}$ and ${\displaystyle b_{i}\,}$ have positive integer values that increase as ${\displaystyle i\,}$ increases to infinity

does this imply that ${\displaystyle x\,}$ is irrational? Frank M Jackson (talk) 17:44, 6 November 2008 (UTC)

Have now found references to Lambert's Irrationality proofs for generalized continued fractions and have added a section with appropriate references. Frank M Jackson (talk) 11:22, 8 November 2008 (UTC)

## Notation: suggest changing to short expression in place of long

Since, to quote the section on Notation:

"The long continued fraction expression ... takes up a lot of space in a book (and it's not easy for the typesetter, either)"

would it not be better to use the short expression

${\displaystyle x=b_{0}+{\frac {a_{1}}{b_{1}+}}{\frac {a_{2}}{b_{2}+}}{\frac {a_{3}}{b_{3}+}}{\frac {a_{4}}{b_{4}+}}\cdots \,}$

in place of the long one

${\displaystyle x=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{b_{4}+{\ddots \,}}}}}}}}}}$

after the introduction?

Here's how the Examples section would appear after the suggested change:

###### ============================================================

Here are two continued fractions that can be built via Euler's identity.

${\displaystyle \log(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots ={\frac {x}{1+}}{\frac {1^{2}x}{2-x+}}{\frac {2^{2}x}{3-2x}}+{\frac {3^{2}x}{4-3x}}+{\frac {4^{2}x}{5-4x}}\cdots \,}$
${\displaystyle \ e^{x}=1+x+{\frac {x^{2}}{2!}}+\cdots =1+{\frac {x}{1-}}{\frac {x}{x+2-}}{\frac {2x}{x+3-}}{\frac {3x}{x+4-}}{\frac {4x}{x+5-}}{\cfrac {5x}{x+6-}}\cdots }$

More advanced techniques are necessary to construct the following examples.

${\displaystyle \ e^{2m/n}=1+{\frac {2m}{n-m+}}{\frac {m^{2}}{3n+}}{\frac {m^{2}}{5n+}}{\frac {m^{2}}{7n+}}{\frac {m^{2}}{9n+}}{\frac {m^{2}}{11n+}}\cdots }$

Setting m = x and n = 2 yields

${\displaystyle \ e^{x}=1+{\frac {2x}{2-x+}}{\frac {x^{2}}{6+}}{\frac {x^{2}}{10+}}{\frac {x^{2}}{14+}}{\frac {x^{2}}{18+}}{\frac {x^{2}}{22+}}\cdots }$
${\displaystyle \pi =3+{\frac {1}{6+}}{\frac {9}{6+}}{\frac {25}{6+}}{\frac {49}{6+}}{\frac {81}{6+}}{\frac {121}{6+}}\cdots }$
${\displaystyle \pi ={\frac {4}{1+}}{\frac {1}{3+}}{\frac {4}{5+}}{\frac {9}{7+}}{\frac {16}{9+}}{\frac {25}{11+}}{\frac {36}{13+}}{\frac {49}{15+}}\cdots }$

Glenn L (talk) 03:06, 5 March 2009 (UTC)

Sounds good to me. Robinh (talk) 08:04, 5 March 2009 (UTC)

## Convergence of Nth roots

Looking at them I don't think the Nth root continued fractions converge. The partial numerators go up as squares whereas the denominators go up linearly. It doesn't seem like a good combination to me. Dmcq (talk) 17:15, 21 November 2009 (UTC)

Oh sorry I see, it's like the expansion for π just above. I'll have to work out for myself what it actually means in convergence terms. Dmcq (talk) 17:28, 21 November 2009 (UTC)
The secret is choosing x and y so that xn > |y|, if possible. Using Pogson's ratio, for example,
${\displaystyle {\sqrt[{5}]{100}}={\sqrt[{5}]{2.5^{5}+{\frac {75}{32}}}}={\sqrt[{5}]{2^{5}+68}},}$
but the first expression converges much more quickly since 97.65625 >> 2.34375, while 32 < 68.
Nevertheless, the second expression does converge. It just takes longer.---Glenn L (talk) 03:21, 22 November 2009 (UTC)