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Iterated continued fraction from convergents are described at https://oeis.org/wiki/Convergents_constant and https://oeis.org/wiki/Table_of_convergents_constants.

Do you think there is any error in the computations, or perhaps in my interpretation, in my new notebook which is pasted to the bottom of this question? It indicates the iterated continued fraction from convergents, or convergents constant (cc), for Pi/2 is 3/2. Can anyone help me prove that Mathematica is correct? The warning statement,"ContinuedFraction::incomp: Warning: ContinuedFraction terminated before 30 terms" makes me worry that there could be some error in the computation.

N[Pi/2,40]

1.570796326794896619231321691639751442099

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.399437073110430452143756644740611223113

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.490343396341538312190367098091859766573

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499506141544753996245023837037133508709

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499975256017086378103094352668241530238

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499998762644884859572966908242122674513

Convergents[%,30];

ContinuedFraction::incomp: Warning: ContinuedFraction terminated before 30 terms. 

N[FromContinuedFraction[%],40]

1.499999938132076351161240652364859960316

Convergents[%,30];

ContinuedFraction::incomp: Warning: ContinuedFraction terminated before 30 terms. 

N[FromContinuedFraction[%],40]

1.499999996906567523984731477342040062854

Convergents[%,30];

ContinuedFraction::incomp: Warning: ContinuedFraction terminated before 30 terms. 

N[FromContinuedFraction[%],40]

1.499999999845328929138608048920727157541

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499999999992275328240194538348680646578

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499999999999614321523514259986324767680

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499999999999978495630126428554334079425

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499999999999430490592898180226181726996

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499999999990876577511876363531307127180

Convergents[%,30];

N[FromContinuedFraction[%],40]

1.499999999999543690097706342514153778772

l=Pi/2; Table[c=Convergents[l,500];l=FromContinuedFraction[c],{a,200}];N[l,300]

1.499999999999999999999999999999999999999999999999999999999999999999999999999999 99999999999999999999999999999999999999999999999999999999999999999999999999999999 99999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999997535967131264972798688959975537466275259

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So I don't know much about this, but is it possible that this process always converges to $3/2$, or at least does so for starting points in a fairly wide range? That seems more likely than that this is some special property of $\pi/2$. –  Michael Lugo Jun 20 '11 at 23:58
    
I agree, I can't prove it, but by trying various starting points, I think it is true for 1<x<2. I used π/2 for an example because it is transcendental. Then there are the open questions, "After how many iterations does the output cease to be transcendental; when does it cease to be normal(If π/2 is normal to begin with) ; finally, when does it cease to be irrational?" –  Marvin Ray Burns Jun 21 '11 at 1:45
1  
Why do you think that the number ever ceases to be irrational or normal? It's probably always (from the second iteration on, if you start with an irrational) transcendental and normal. The fact that it is close to $3/2$ doesn't mean that it's not normal - the error term is "random" and therefore probably normal. –  Yuval Filmus Jun 21 '11 at 8:00

1 Answer 1

up vote 9 down vote accepted

Suppose that the original (irrational) number has the continued fraction $[1;x,y,\ldots]$. After one iteration, the number is of the form $$1+\frac{1}{1+1/x+\frac{1}{1+1/(x+1/y)+\epsilon}},$$ where $\epsilon < 1$. Its continued fraction starts $[1;z,w,\ldots]$, where $z$ is the floor of $$1+1/x+\frac{1}{1+1/(x+1/y)+\epsilon}.$$ It's easy to see that this quantity is always less than $3$, so either $z=1$ or $z=2$. If $x=1$ then clearly $z=2$. Therefore after at most two iterations we reach a number with $z=2$. In this case $w$ is the floor of $$ \begin{align*} &\frac{1}{\frac{1}{1+1/(2+1/y)+\epsilon} - \frac{1}{2}} \\ = &\frac{1}{\frac{2+1/y}{3+1/y+(2+1/y)\epsilon} - \frac{1}{2}} \\ = &\frac{6+2/y+(4+2/y)\epsilon}{1+1/y-(2+1/y)\epsilon} \\ = &\frac{6y+2+(4y+2)\epsilon}{y+1-(2y+1)\epsilon}. \end{align*} $$ Denote by $r = [1;2,y,\ldots]$ the original number. So $$ r = [1;2,y,\ldots] \geq 1+\frac{1}{2+1/y} = \frac{3y+1}{2y+1}.$$ Denote the convergents of $r$ by $r_1=1,r_2=3/2,r_3,\ldots$. For $i \geq 3$ we have $$\frac{3y+1}{2y+1} \leq r \leq \frac{3}{2}.$$ Denote these by $r_l < r < r_h$. Therefore $$ \begin{align*} \epsilon &= [0;r_3,r_4,\ldots] \\ &\geq [0;r_h,r_l,r_h,r_l,\ldots] \\ &= \frac{\sqrt{r_l^2 r_h^2 + 4r_l r_h} - r_lr_h}{2r_l} \geq \frac{1}{2} - \frac{1}{60y}. \end{align*} $$ The last inequality can be obtained through laborious Taylor expansion. Substituting this value in the lower bound for $w$, we get that it is at least $$\frac{312y^2+126y}{32y+1} \geq \frac{312}{32} y. $$ Therefore the third coefficient of the continued fraction increases exponentially. We deduce that the iterates converge exponentially to $[1;2] = 3/2$.

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