In today's blog, I will show proofs for two properties of Continued Fractions (for those not familiar with Continued Fractions, start here):

- All rational numbers can be represented as a finite continued fraction.
- All irrational numbers that are solutions to a quadratic equation are periodic (that is, they repeat the same pattern of integers over and over again after a certain point)

ax

^{2}+ bx + c = 0

where a,b,c are integers.

For a periodic continued fraction which a period of size j and where the period begins before i, then: a

_{i}= a

_{i+j}.

The proofs outlined in today's blog are based on work done by Harold M. Stark in his book An Introduction to Number Theory.

Theorem 1: A continued fraction is finite if and only if it is a rational number.

(1) We know that if a continued fraction is finite, then it is representable as a rational number. (See Lemma 2, here)

(2) So, all we need to prove is that a rational number is representable as a finite continued fraction.

(3) For any rational number, there are two integers, let's call them c,d

_{0}such that the rational number is representable by a/b. [Definition of a rational number]

(4) To generate a continued fraction, we need to generate a set of integers a

_{0}through a

_{n}.

(5) Let's set a

_{0}= floor(c/d

_{0}).

A floor is a function that returns the minium integer that is less than or equal to a certain rational value. For example, the floor of (3/4) is 0. The floor of (5/2) is 2 and the floor of (-8) is -8. Finally, the floor of (-8.1) is -9.

(6) If a

_{0}= c/d

_{0}, then we are done and the continued fraction is [a

_{0}]. Going forward, we will assume that a

_{0}is less than but not equal to c/d

_{0}.

(7) Subtracting the original number by the a

_{0}results in a rational number which we can call β

_{0}:

c/d

_{0}= a

_{0}+ β

_{0}

We note that β

_{0}is ≥ 0 and less than 1 by our assumption in #6. When it is 0 we are done.

(8) Now, from #7, we know that d

_{0}* β

_{0}is an integer since:

c = d

_{0}* a

_{0}+ d

_{0}*β

_{0}

And therefore:

d

_{0}* β

_{0}= c - d

_{0}*a

_{0}

Which is an integer since c is an integer and d

_{0}is an integer and a

_{0}is an integer.

(9) So, we know there exists an integer d

_{1}= d

_{0}* β

_{0}.

We also note that this integer has the following properties:

(a) d

_{1}≥ 0.

We know that it is not negative since we can assume d

_{0}is positive (if both c,d

_{0}are negative, we can assume that they are both positive. If c is positive and d

_{0}is negative, we can assume that c is negative and d

_{0}is positive.

(b) d

_{1}is less than d

_{0}

We know it is less than d

_{0}since β

_{0}is less than 1 and since d

_{1}= d

_{0}* β

_{0}

(10) From #9, we note that β

_{0}= d

_{1}/d

_{0}

(11) Now, a

_{1}= floor(1/β

_{0}) = floor(d

_{1}/d

_{0})

Now if this is an integer (for example, 1/(1/2) = 2), then we are done and the continued fraction is [a

_{0},a

_{1}]. So, let's assume that we are not done.

(12) So, then, there exists rational value β

_{1}that is equal to the difference between 1/β

_{0}and a

_{1}which is greater than 0 so that:

d

_{1}/d

_{0}= a

_{1}+ β

_{1}

This is the same form as step #7 and the same principles apply. We can derive a d

_{2}that is a positive integer less than d

_{1}and so on.

(13) Because d

_{i}is an integer that continually gets smaller, we know that eventually it will equal 0.

(14) When d

_{i}= 0 we are done and so we have proven that our sequence will eventually end.

QED

Lemma 1: If a continued fraction is periodic, then it represents a quadratic number.

(1) A quadratic number is any real number that satisfies the quadratic equation:

ax

^{2}+ bx + c = 0 where a,b,c are integers.

For example, √2 is a quadratic number since it satisfies the equation (a=1,b=0,c=-2) where:

x

^{2}- 2 = 0

(2) Let α be a real number that is represented by a periodic continued fraction that repeats a a

_{m}with a period of k.

So α = [ a

_{0}, a

_{1}, ..., a

_{m-1}, α

_{m}]

And

α

_{m}= [ a

_{m}, a

_{m+1}, ..., a

_{m+k-1}, a

_{m}, a

_{m+1}, ...]

And

α

_{m}= [ a

_{m}, a

_{m+1}, ..., a

_{m+k-1}, α

_{m}]

(3) So, applying Lemma 1 from a previous blog, we get:

α

_{m}= (α

_{m}p

_{m}+ p

_{m-1})/(α

_{m}q

_{m}+ q

_{m-1})

(4) And multiplying (α

_{m}q

_{m}+ q

_{m-1}) to both sides gives us:

(α

_{m})

^{2}*q

_{m}+ α

_{m}*q

_{m-1}= (α

_{m}p

_{m}+ p

_{m-1})

And:

(α

_{m})

^{2}*q

_{m}+ α

_{m}*q

_{m-1}- α

_{m}p

_{m}- p

_{m-1 }= 0

And:

(α

_{m})

^{2}*q

_{m}+ (α

_{m})(q

_{m-1}- p

_{m}) - p

_{m-1}= 0

(5) Applying lemma 1 from a previous blog to the first equation in #2 gives us:

α = (α

_{m}*p

_{m-1}+ p

_{m-2})/(α

_{m}*q

_{m-1}+ q

_{m-2})

(6) So,

α*(α

_{m}*q

_{m-1}+ q

_{m-2}) = (α

_{m}*p

_{m-1}+ p

_{m-2})

and

α*α

_{m}*q

_{m-1}+ α*q

_{m-2}- α

_{m}*p

_{m-1}- p

_{m-2}= 0

and

α

_{m}(α*q

_{m-1}- p

_{m-1}) = p

_{m-2}- α*q

_{m-2}

and

α

_{m}= (p

_{m-2}- α*q

_{m-2})/(α*q

_{m-1}- p

_{m-1})

(7) Now inserting (#6) into (#4) gives us:

[ ([p

_{m-2}- α*q

_{m-2})/(α*q

_{m-1}- p

_{m-1})]

^{2}*q

_{m}+

[(p

_{m-2}- α*q

_{m-2})/(α*q

_{m-1}- p

_{m-1})](q

_{m-1}- p

_{m}) - p

_{m-1}= 0

And multiplying (α*q

_{m-1}- p

_{m-1})

^{2}to both sides give us:

(p

_{m-2}- α*q

_{m-2})

^{2}*q

_{m}+

(p

_{m-2}- α*q

_{m-2})(α*q

_{m-1}- p

_{m-1})(q

_{m-1}- p

_{m}) - (α*q

_{m-1}- p

_{m-1})

^{2}(p

_{m-1}) = 0

(8) And solving for all the values above gives us a quadratic equation of the form:

α

^{2}a + αb + c = 0

QED

Lemma 2: If α is a quadratic number, and [a

_{0}, a

_{1}, ... , a

_{n-1}, α

_{n}] is a continued fraction representation where a

_{i}are integers and α

_{n}is a real number, then α

_{n}is also a quadratic number.

(1) Since α is a quadratic number, there exists integers a,b, and c such that:

a(α)

^{2}+ b(α) + c = 0. [Definition of a quadratic number]

(2) From Lemma 1 in a previous blog, we know that:

α = (α

_{n}* p

_{n-1}+ p

_{n-2})/(α

_{n}* q

_{n-1}+ q

_{n-2})

(3) Inserting (1) into (2) and then multiplying everything out and then regrouping around (α

_{n})

^{2}, α

_{n}, and any integer gives us three values A

_{n}, B

_{n}, C

_{n}such that:

A

_{n}(α

_{n})

^{2}+ B

_{n}(α

_{n}) + C

_{n}= 0.

where

A

_{n}= a(p

_{n-1})

^{2}+ bp

_{n-1}q

_{n-1}+ c(q

_{n-1})

^{2}

B

_{n}= 2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}

C

_{n}= a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}= A

_{n-1}

(4) By definition, this means that α

_{n}is a quadratic number.

QED

Corollary 2.1: (B

_{n})

^{2}- 4A

_{n}C

_{n}= b

^{2}- 4ac

where:

A

_{n}= a(p

_{n-1})

^{2}+ bp

_{n-1}q

_{n-1}+ c(q

_{n-1})

^{2}

B

_{n}= 2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}

C

_{n}= a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}= A

_{n-1}

(1) From Lemma 2 above:

(B

_{n})

^{2}- 4A

_{n}C

_{n}=

= (2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) - 4*[a(p

_{n-1})

^{2}+ bp

_{n-1}q

_{n-1}+ c(q

_{n-1})

^{2}][a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}= A

_{n-1}] =

= (2ap

_{n-1}p

_{n-2})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) +

(bp

_{n-1}q

_{n-2})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) +

(bp

_{n-2}q

_{n-1})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) +

(2cq

_{n-1}q

_{n-2})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) -

(4a(p

_{n-1})

^{2})[a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}] -

(4bp

_{n-1}q

_{n-1})[a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}] -

(4c(q

_{n-1})

^{2})[a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}]

(2) Multipyling each term, we get:

(i) (2ap

_{n-1}p

_{n-2})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) =

4a

^{2}(p

_{n-1})

^{2}(p

_{n-2})

^{2}+

2ab(p

_{n-1})

^{2}(p

_{n-2})(q

_{n-2}) +

2ab(p

_{n-1})(p

_{n-2})

^{2}(q

_{n-1}) +

4ac(p

_{n-1})(p

_{n-2})(q

_{n-1})(q

_{n-2})

(ii) (bp

_{n-1}q

_{n-2})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) =

2ab(p

^{n-1})

^{2}(p

_{n-2})(q

_{n-2}) +

b

^{2}(p

_{n-1})

^{2}(q

_{n-2})

^{2}+

b

^{2}(p

_{n-1})(p

_{n-2})(q

_{n-1})(q

_{n-2}) +

2bc(p

_{n-1})(q

_{n-1})(q

_{n-2})

^{2}

(iii) (bp

_{n-2}q

_{n-1})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) =

2ab(p

_{n-1})(p

_{n-2})

^{2}(q

_{n-1}) +

b

^{2}(p

_{n-1})(p

_{n-2})(q

_{n-1})(q

_{n-2}) +

b

^{2}(p

_{n-2})

^{2}(q

_{n-1})

^{2}+

2bc(p

_{n-2})(q

_{n-1})

^{2}(q

_{n-2})

(iv) (2cq

_{n-1}q

_{n-2})(2ap

_{n-1}p

_{n-2}+ bp

_{n-1}q

_{n-2}+ bp

_{n-2}q

_{n-1}+ 2cq

_{n-1}q

_{n-2}) =

4ac(p

_{n-1})(p

_{n-2})(q

_{n-1})(q

_{n-2}) +

2bc(p

_{n-1})(q

_{n-1})(q

_{n-2})

^{2}+

2bc(p

_{n-2})(q

_{n-1})

^{2}(q

_{n-2}) +

4c

^{2}(q

_{n-1})

^{2}(q

_{n-2})

^{2}

(v) -(4a(p

_{n-1})

^{2})[a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}

_{ }] =

-4a

^{2}(p

_{n-1})

^{2}(p

_{n-2})

^{2}+ -4ab(p

_{n-1})

^{2}(p

_{n-2})(q

_{n-2}) + -4ac(p

_{n-1})

^{2}(q

_{n-2})

^{2}

(vi) -(4bp

_{n-1}q

_{n-1})[a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}] =

-4ab(p

_{n-1})(p

_{n-2})

^{2}(q

_{n-1}) +

-4b

^{2}(p

_{n-1})(p

_{n-2})(q

_{n-1})(q

_{n-2}) +

-4bc(p

_{n-1})(q

_{n-1})(q

_{n-2})

^{2}

(vii) -(4c(q

_{n-1})

^{2})[a(p

_{n-2})

^{2}+ bp

_{n-2}q

_{n-2}+ c(q

_{n-2})

^{2}] =

-4ac(p

_{n-2})

^{2}(q

_{n-1})

^{2}+

-4bc(p

_{n-2})(q

_{n-1})

^{2}(q

_{n-2}) +

-4c

^{2}(q

_{n-1})

^{2}(q

_{n-2})

^{2}

(3) We can see that many of these terms line up and cancel out:

= [ 4a

^{2}(p

_{n-1})

^{2}(p

_{n-2})

^{2}- 4a

^{2}(p

_{n-1})

^{2}(p

_{n-2})

^{2}]+ { from 2.i and 2.v }

[ 2ab(p

_{n-1})

^{2}p

_{n-2}q

_{n-2}+ 2ab(p

_{n-1})

^{2}p

_{n-2}q

_{n-2}- 4ab(p

_{n-1})

^{2}p

_{n-2}q

_{n-2 }] + { from 2.i, 2.ii, and 2.iv }

[ 2ab(p

_{n-1})(p

_{n-2})

^{2}(q

_{n-1}) + 2ab(p

_{n-1})(p

_{n-2})

^{2}(q

_{n-1}) - 4ab(p

_{n-1)}(p

_{n-2})

^{2}(q

_{n-1}) ]+ { from 2.i, 2.iii, and 2.vi }

[ 2bcp

_{n-1}(q

_{n-1})

^{2}q

_{n-2}+ 2bcp

_{n-1}(q

_{n-1})

^{2}q

_{n-2 }- 4bcp

_{n-1}(q

_{n-1})

^{2}q

_{n-2}] + { from 2.iii, 2.iv, and 2.vii }

[ 2bcp

_{n-2}q

_{n-1}(q

_{n-2})

^{2}+2bcp

_{n-2}q

_{n-1}(q

_{n-2})

^{2}- 4bcp

_{n-2}q

_{n-1}(q

_{n-2})

^{2}] + { from 2.ii, 2.iv, and 2.vi }

( 4c

^{2}(q

_{n-1})

^{2}(q

_{n-2})

^{2}- 4c

^{2}(q

_{n-1})

^{2}(q

_{n-2})

^{2}) + { from 2.iv and 2.vii }

( b

^{2}p

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}+ b

^{2}p

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2 }- 4b

^{2}p

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}) + { from 2.ii, 2.iii, and 2.vi }

(4acp

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}+ 4acp

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}) + { from 2.i and 2.iv }

[b

^{2}(p

_{n-1})

^{2}(q

_{n-2})

^{2}+ b

^{2}(p

_{n-2})

^{2}(q

_{n-1})

^{2}] + { from 2.ii and 2.iii }

[-4ac(p

_{n-2})

^{2}(q

_{n-1})

^{2}+-4ac(p

_{n-1})

^{2}(q

_{n-2})

^{2}] { from 2.v and 2.vii }

= -2b

^{2}p

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2 + }8acp

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}+

b

^{2}(p

_{n-1})

^{2}(q

_{n-2})

^{2}+

b

^{2}(p

_{n-2})

^{2}(q

_{n-1})

^{2}+

-4ac(p

_{n-2})

^{2}(q

_{n-1})

^{2}+

-4ac(p

_{n-1})

^{2}(q

_{n-2})

^{2}=

= b

^{2}[(p

_{n-1})

^{2}(q

_{n-2})

^{2}+ (p

_{n-2})

^{2}(q

_{n-1})

^{2}- 2p

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}] - 4ac[(p

_{n-1})

^{2}(q

_{n-2})

^{2}+ (p

_{n-2})

^{2}(q

_{n-1})

^{2}- 2p

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}] =

= (b

^{2}- 4ac)[(p

_{n-1})

^{2}(q

_{n-2})

^{2}+ (p

_{n-2})

^{2}(q

_{n-1})

^{2}- 2p

_{n-1}p

_{n-2}q

_{n-1}q

_{n-2}] =

= (b

^{2}- 4ac)(p

_{n-1}q

_{n-2}- p

_{n-2}q

_{n-1})

^{2}

(4) Applying Lemma 2 from a previous blog, we get:

(b

^{2}- 4ac)(p

_{n-1}q

_{n-2}- p

_{n-2}q

_{n-1})

^{2}= (b

^{2}- 4ac)(-1)

^{2}=

= b

^{2}- 4ac.

QED

Corollary 2.2: A

_{n}, B

_{n}, C

_{n}can only span over a finite set of integers.

(1) Let λ

_{n}= p

_{n}q

_{n}- α(q

_{n})

^{2}

(2) Rearranging values gives us:

p

_{n}= λ

_{n}/q

_{n}- αq

_{n}

(3) We note that absolute(λ

_{n}) is less than 1 since:

(a) We know that α - p

_{n}/q

_{n}is less than 1/q

_{n}

^{2}and greater than -1/q

_{n}

^{2}. [From Theorem 2, here]

(b) Multiplying (q

_{n})

^{2}to all sides give us that:

α*(q

_{n})

^{2}- p

_{n}q

_{n}is between -1 and 1.

(c) Multiplying -1 to all sides give us that:

p

_{n}q

_{n}- α(q

_{n})

^{2}is between -1 and 1.

(d) This proves that λ

_{n}is between -1 and 1 or rather absolute(λ

_{n}) is less than 1.

(4) Inserting #2 into the formula for A

_{n}in Lemma 2 gives us:

A

_{n}= a( λ

_{n-1}/q

_{n-1}+ αq

_{n-1})

^{2}+ b( λ

_{n-1}/q

_{n-1}+ αq

_{n-1})a

_{n-1}+ c(q

_{n-1})

^{2}

Working this through and rearranging values gives us:

(q

_{n-1})

^{2}(aα

^{2}+ bα + c) + 2aαλ

_{n-1}+ a(λ

_{n-1})

^{2}/(q

_{n-1})

^{2}+ bλ

_{n-1}

Since we know that aα

^{2}+ bα + c = 0 (since α is a quadratic number based on a,b,c), we get:

A

_{n}= 2aαλ

_{n-1}+ a(λ

_{n-1})

^{2}/(q

_{n-1})

^{2}+ bλ

_{n-1}

Since absolute(λa

_{n-1}) is less than 1, this tells us that:

absolute(A

_{n}) is less than 2*absolute(aα) + absolute(a) + absolute(b).

To understand the details for why this is true, there are three cases to consider:

Case I: a,b positive

In this case A

_{n}is less than 2*absolute(aα) + absolute(a) + absolute(b).

Case II: a,b negative

In this case, the absolute values result in values the same as in Case I. In this case A

_{n}= -absolute(A

_{n}).

Case III: a,b: one is negative, one is positive

In this case, A

_{n}will be greater than -absolute(A

_{n}) and less than absolute(A

_{n}) so, then:

absolute(A

_{n}) is less than 2*absolute(aα) + absolute(a) + absolute(b)

Since C

_{n}= A

_{n-1}, we know that:

absolute(C

_{n}) is less than 2*absolute(aα) + absolute(a) + absolute(b)

Finally, since (B

_{n})

^{2}= 4A

_{n}C

_{n}+ b

^{2}-4ac, we know that:

(B

_{n})

^{2}≤ 4[ 2*absolute(aα) + absolute(a) + absolute(b)]

^{2}+ absolute(b

^{2}-4ac)

(5) The important idea from all this is that there are only a finite number integers that A

_{n}, B

_{n}, and C

_{n}can equal, that is, only those integers which make up the ranges in #4.

QED

Theorem 2: A continued fraction is periodic if and only if it represents an irrational quadratic number.

(1) In Lemma 1, it was shown that a periodic continued fraction is a quadratic irrational number.

(2) Now, I will show that all quadratic irrational numbers are representable as periodic continued fractions.

(3) Let α be a quadratic number (that is, an irrational that is a solution to a quadratic equation)

(4) We know that there are an infinite number of values a

_{i}and α

_{i}that make it up (from Theorem 1 above, otherwise it would not be irrational)

(5) But at each point, there are only a finite number of values A

_{n}, B

_{n}, C

_{n}that describe the equation that α

_{i}solves.

(6) This means that it is inevitable that a specific combination A

_{n}, B

_{n}, C

_{n}repeats and continues to repeat.

(7) For each specific combination of A

_{n}, B

_{n}, C

_{n}, there are two possible real values that satisfy the equation. (This observation comes from the solution to the quadratic equation, found here)

(8) So, eventually, some α

_{i}repeats.

QED

Lemma 3: If a value is representeable by a finite continued fraction with an odd number of elements, then it is representable by an even number of elements and likewise if it is representable by an even number of elements, it can be represented by an odd number of elements.

(1) Let [a

_{0}, a

_{1}, ..., a

_{n}] be a finite continued fraction of n elements.

(2) Now, if a

_{n}≥ 2, then:

[ a

_{0}, a

_{1}, ..., a

_{n}] = [ a

_{0}, a

_{1}, ..., a

_{n}-1,1 ]

(3) Otherwise, if a

_{n}= 1, then:

[ a

_{0}, a

_{1}, a

_{n-1},1 ] = [a

_{0}, a

_{1}, ... , a

_{n-1}+ 1 ]

QED

## 8 comments:

In Lemma 1 (4):

Should it be:

(αm)^2*qm + αm*qm-1 = (αmpm + pm-1)

Instead of:

(αm)^2*pm + αm*pm-1 = (αmqm + qm-1) ?

Same in Lemma 1 (6).

In Corollary 2.1:

(fourth line of equality):

Should be:

4ab(pn-1)(pn-2)^2(qn-1)

Instead of:

4ab(pn-1)(pn-2)^2(qn-2) ?

(fifth line of equality):

Should be:

4bc(pn-1)(qn-1)(qn-2)^2

Instead of:

4bc(pn-1)(qn-1)^2(qn-2) ?

(sixth line of equality):

Should be:

4bc(pn-2)(qn-1)^2(qn-2)

Instead of:

4bc(pn-2)(qn-1)(qn-2)^2 ?

In Corollary 2.2 (4):

b(λn/qn + αqn-1)an-1

Should be:

b(λn-1/qn-1 + αqn-1)qn-1 ?

Hi Rob,

Awesome job in reconstructing the correct argument. In looking the proof (2.1) over, I realized that it needed to be reworked.

I added additional details in an effort to make the argument clear and fix mistakes.

Thanks so much for your comments! :-)

Regards,

-Larry

In step (1) of

Theorem 1.Should the reference be

Theorem 1instead ofLemma 2?Rob

In step (3) of

Theorem 1should:a/b

be

c/d0

Rob

In step (3) of

Lemma 1should:p(m),q(m),p(m-1),q(m-1)

be replaced with:

p(m-k-1),q(m-k-1),p(m-k-2),q(m-k-2)

Which then assumes that k>1. Although I suppose you can substitute:

αm=[am,αm]

with

αm=[am,am+1,αm]

and then have am+1=am so k=2.

I am finding these continued fractions difficult so if I have got it all wrong then please let me know.

Thanks

Rob

In step (2) of

Corollary 2.2:should

pn=λn/qn-αqn

be

pn=λn/qn

+αqnIn step (4) of

Corollary 2.2:should

An=a(λn-1/qn-1+αqn-1)^2+b(λn-1/qn-1 +αqn-1)an-1+c(qn-1)^2

be

An=a(λn-1/qn-1+αqn-1)^2+b(λn-1/qn-1 +αqn-1)

n-1+c(qn-1)^2qRob

In step (4) of

Corollary 2.2:should

Since absolute(λan-1) is less than 1

be

Since absolute(λn-1) is less than 1

Rob

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