Talk:Continued fraction

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[edit] citation needed

"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)

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)

[edit] Rational induction

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)

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)

[edit] representation of 1

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)

[edit] Theorem 2

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)

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)
THANK YOU!!!! TechnoFaye Kane 20:46, 8 February 2008 (UTC)

[edit] infinite continued fraction

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)
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)