Why are the continued fractions of non-square-root numbers ($\sqrt[a]{x}$ where $a>2$) not periodic?

Question

Ok, so it is quite amazing how the continued fractions for $\sqrt[2]{x}$ are always periodic for all whole numbers of $x$ (and where $x$ is not a perfect square): Here is a link I suggest at looking at: http://mathworld.wolfram.com/PeriodicContinuedFraction.html ...

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The following is in the simple continued fraction form: $$\sqrt x = a_1+ \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \cfrac{1}{a_5 + \cfrac{1}{a_6 + \cfrac{1}{a_7 + \cdots}}}}}}$$

For example:

The continued fraction of $\sqrt 7$ is as follows: $[2;1,1,1,4,1,1,1,4,1,1,1,4,\ldots]$ the period in this case is $4$... and the continued fraction can be written as:

$$[2;\overline{1,1,1,4}]$$

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Some other examples:

$\sqrt2 = [1;\overline{2}]$ Period is 1

$\sqrt3 = [1;\overline{1,2}]$ Period is 2

$\sqrt{13} = [3;\overline{1,1,1,1,6}]$ Period is 5

$\sqrt{97} = [9;\overline{1,5,1,1,1,1,1,1,5,1,18}]$ Period is 11

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There are many other sources which show this... but why does this not work for others, such as cubic roots? I have written a java program to compute the continued fraction for the nth root of the numbers between $1$ and $100$ (excluding perfect squares,etc...) for the first 100 terms. Here are the results:

$\sqrt[2]{x}$: http://pastebin.com/ZcasfRyP

$\sqrt[3]{x}$: http://pastebin.com/XG9UF8hR

$\sqrt[4]{x}$: http://pastebin.com/Edp307SE

$\sqrt[5]{x}$: http://pastebin.com/9SwwPqUa

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As you can see no period...

So why is it periodic for square roots, but not for others? An extension of the question: https://math.stackexchange.com/questions/1898902/periodic-continued-fractions-of-non-square-root-numbers-sqrtax-where-a

Kind Regards

Joshua Lochner

Answer 1

The easiest way to show this is the contrapositive: if a continued fraction is periodic then its value is a quadratic.

To see this, note that if we have $y = [a_1; a_2, a_3, \ldots, a_n, x]$ for reals $x,y$ and integers $a_i$, then we can write $y=\dfrac{mx+n}{px+q}$ for some integers $m, n, p, q$. (This is easy to prove by induction - note that if $z=[a_2; a_3, \ldots, a_n, x]$ $= \dfrac {m'x+n'}{p'x+q'}$, then $y=a_1+\dfrac1z$ $=a_1+\dfrac{p'x+q'}{m'x+n'}$ $=\dfrac{a_1m'x+a_1n'+p'x+q'}{m'x+n'}$ $=\dfrac{(a_1m'+p')x+(a_1n'+q')}{m'x+n'}$ is clearly of the required form.)

Now, if the continued fraction for $x$ is periodic then we have $x=[a_1; a_2, a_3, \ldots, a_n, x]$, so by the above there are integers with $x=\dfrac{mx+n}{px+q}$. But now multiply both sides by $px+q$ to get $px^2+qx=mx+n$, or $px^2+(q-m)x-n=0$, so that $x$ is a solution to this quadratic equation.

To answer the question about patterns in the continued fractions of other numbers: to the best of my knowledge, nothing is known about the continued fractions of e.g. cube roots — not even whether their coefficients are bounded! — though it's known that they can't grow too quickly: this is a corollary of Roth's Theorem, which bounds the so-called irrationality measure of algebraic numbers (how well they can be approximated by rationals). On the other hand, there is a special class of (transcendental) numbers for which more information on the continued fraction representation is known: certain Liouville numbers have continued fractions with complicated recursive (almost fractal) structures; see http://mathworld.wolfram.com/LiouvillesConstant.html and some of the references there for a little more information on this.

Answer 2

it does not work because, given a strictly periodic continued fraction, we can construct the 2 by 2 matrix that comes from it. Then we get the integer quadratic form that goes with that automorphism matrix. Finally, we get the "quadratic irrational" larger than one that goes with that.

Alright, we are dealing with an indefinite binary quadratic form, $$ f(x,y) = A x^2 + B xy + C y^2, $$ with positive non-square discriminant $$ \Delta = B^2 - 4AC. $$ We find the minimum solution $(\tau, \sigma)$ to the Pell type equation $$ \tau^2 - \Delta \sigma^2 = 4. $$ Then the matrix that generates the ($+1$) automorphism group of the form is $$ P = \left( \begin{array}{rr} \frac{\tau - B \sigma}{2} & -C \sigma \\ A \sigma & \frac{\tau + B \sigma}{2} \end{array} \right) $$

The Hessian matrix of the form is $$ H = \left( \begin{array}{rr} 2 A & B \\ B & 2C \end{array} \right) $$ and the automorphism matrices satisfy $$ P^T H P = H. $$ In the discussion below, the "quadratic irrational" that has that continued fraction comes from the quadratic formula going right to left, that is $$ \frac{-13 - \sqrt {345}}{-22} \approx 1.4351898 $$

Next, a CF that is "eventually periodic" is just a rearrangement of the quadratic irrational. Let me give a "reduced" quadratic form as example, give me a minute

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 4 13 -11

  0  form              4          13         -11


           1           0
           0           1

To Return  
           1           0
           0           1

0  form   4 13 -11   delta  -1
1  form   -11 9 6   delta  2
2  form   6 15 -5   delta  -3
3  form   -5 15 6   delta  2     ambiguous  
4  form   6 9 -11   delta  -1
5  form   -11 13 4   delta  3
6  form   4 11 -14   delta  -1
7  form   -14 17 1   delta  17
8  form   1 17 -14   delta  -1     ambiguous  
9  form   -14 11 4   delta  3
10  form   4 13 -11


  form   4 x^2  + 13 x y  -11 y^2 

minimum was   1rep   x = 108   y = 155 disc 345 dSqrt 18  M_Ratio  20.25
Automorph, written on right of Gram matrix:  
-2029  -8008
-2912  -11493
=========================================

Given the matrix I call "Automorph," here is how we recover the form:

? m = [ 2029, 8008; 2912, 11493]
%2 = 
[2029 8008]

[2912 11493]

? matdet(m)
%3 = 1
? 2029 + 11493
%4 = 13522
? gcd(5,3)
%5 = 1
? gcd(2912, 8008)
%6 = 728
-
? tr = 2029 + 11493
%7 = 13522
? g = gcd(2912, 8008)
%8 = 728
? tr^2 - 4
%9 = 182844480

? rat = ( tr^2 - 4) / g^2
%12 = 345

? tr^2 - 345 * g^2
%14 = 4
? 
? a = 2912 / g
%16 = 4

? b = ( tr - 2 * 2029) / g
%18 = 13

? c = -8008 / g
%17 = -11

Given the purely periodic continued fraction with repeated part $$ 1,2,3,2,1,3,1,17,1,3, $$ the resulting matrix is the product of little 2 by 2 matrices:

? a1 = [0,1; 1,1]
%1 = 
[0 1]

[1 1]

? a2 = [0,1; 1,2]
%2 = 
[0 1]

[1 2]

? a3 = [0,1; 1,3]
%3 = 
[0 1]

[1 3]

? a17 = [0,1; 1,17]
%4 = 
[0 1]

[1 17]

? auto = a1 * a2 * a3 * a2 * a1 * a3 * a1 * a17 * a1 *a3
%5 = 
[2029 8008]

[2912 11493]

? 

Answer 3

The idea is quite simple. Suppose that $$x=[a_0;\overline{a_1 , \dots , a_k}]$$ is a real number whose continued fraction is periodic. Then it satisfies the equation $$x- a_0= \frac{1}{a_1+\frac{1}{\frac{\dots}{a_{k-1}\frac{1}{a_k+(x-a_0)}}}}$$ which can be transformed by a quadratic equation. with integer coefficients. For example, the number $x=[2; \overline{1,4}]$ satisfies $$x-2= \frac{1}{1+\frac{1}{4+\frac{1}{x-2}}}$$ which is equivalent to the quadratic equation $$(x-2)(5x-9)=4x-7$$

So, all periodic numbers must be quadratic.

Answer 4

It's because periodicity of a simple continued fraction amounts to a quadratic equation. I will illustrate this by means of an example: $$ 7 + \cfrac 1 {2 + \cfrac 1 {3+\cfrac 1 {9 + \cdots\vphantom{\dfrac 1 1}}}} $$ and assume $2,\,3,\,9$ repeats. We have $$ 7,\ \overbrace{2,3,9,}\ \overbrace{2,3,9,}\ \overbrace{2,3,9,}\ \overbrace{2,3,9,}\ \ldots $$ This is $$ -2 + \left(9 + \cfrac 1 {2 + \cfrac 1 {3+\cfrac 1 {9 + \cdots\vphantom{\dfrac 1 1}}}} \right) \tag 1 $$ so that $9,\ 2,\ 3$ repeats right from the beginning.

Let $x = {}$the continued fraction in $(1)$, with $9,\ 2,\ 3$ repeating.

Then we get $$ x = 9 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 x}} $$ Then, since $\dfrac 1 {3+\cfrac 1 x} = \dfrac x {3x+1}$, we have $$ x = 9 + \cfrac 1 {2 + \cfrac x {3x+1}}. $$ Now multiply the numerator and denominator of the fraction after $9+\cdots$ by $3x+1$, getting $$ x = 9 + \dfrac{3x+1}{7x+2} $$ so $$ x = \frac{66x+ 19}{7x+2}. $$ Multiply both sides by $7x+2$: $$ x(7x+2) = 66x+19. $$ And there you have a quadratic equation.