Various Old and New Results in Classical Arithmetic by Special Functions

Home , Fibonacci polynomials, Liouville number

VARIOUS OLD AND NEW RESULTS IN CLASSICAL ARITHMETIC BY SPECIAL FUNCTIONS

A thesis submitted To Kent State University in partial Fulﬁllment of the requirements for the Degree of Master of Science

Michael Henry

May 2018 c Copyright All rights reserved Except for previously published materials Thesis written by Michael Andrew Henry B.A., The University of Maryland College Park, 2009 M.S., Kent State University 2018

Approved by Gang Yu ______, Advisor

Andrew Tonge______, Chair, Department of Mathematical Sciences

James L. Blank______, Dean, College of Arts and Sciences

Table of Contents

Page

List of Tables ...... iv

Acknowledgments ...... v

1. INTRODUCTION ...... 1

2. BASIC NUMBER THEORY RESULTS ...... 2

3. IRRATIONALITY ...... 5

4. TRANSCENDENCE ...... 14

5. ELEMENTARY CONTINUED FRACTIONS ...... 20

6. LAMBERT SERIES, ϑ-FUNCTIONS, AND LANDAU’S RECIPROCAL FIBONACCI SUM...... 30

7. FIBONACCI POLYNOMIALS AND MORE LANDAU SUMS ...... 39

8. LATERAL LANDAU SUMS ...... 44

9. SPECULATIONS AND FINAL REMARKS ...... 47

iii List of Tables

Table Page

5.1 Continuant polynomials ...... 22

7.1 Fibonacci polynomials ...... 42

7.2 Values of qn(k) ...... 43

8.1 A glimpse of the spectrum ...... 46

iv Acknowledgments

I would like to thank Dr. Gang Yu for his assistance in a reading course in number theory that was the basis of this work; I am very grateful for the degree of independence that he has given me throughout my studies. I would like to thank Dr. Ulrike Vorhauer for her encouragement and enthusiasm for me to continue. Furthermore, she has a clarity of thought that is enviable and that has benefited the final form of this thesis tremendously. I have been influenced also by many stimulating conversations with Dr. Omar De la Cruz Cabrera (who is both a statistician and a logician) and Dr. Morley Davidson, the latter having, what I would call, a deeply holistic perspective of mathematics. Also, thanks is due to Dr. Artem Zvavitch, who introduced me to yet another layer of beauty that lies behind mathematical analysis.

This work is dedicated to M.E.T.

v 1: INTRODUCTION

We will prove Dirichlet’s rational approximation theorem, which frames our discussion and

approach to Diophantine approximation. We proceed to prove that Liouville’s number L is transcendental. Then we complement these results with some theory of simple continued fractions and related results. We state the order of approximation of e, which by Roth’s theorem gives us e is a transcendental number. At this point, we enter what is probably best described as the intersection of classical analysis and classical arithmetic, with special attention given to continued fractions. We generalize a result of Landau, giving rise to what we will call Landau sums. These sums generate a countable spectrum of irrational numbers that are computable in quadratic time due to their representation as theta functions; it is still open as to whether these numbers are transcendental. Altering the Landau sums we also get another class of sums that we call lateral Landau sums. These sums also give a spectrum of numbers that we speculate are irrational. These results sit nicely within the intersection of classical analysis and classical number theory. We say a few words about this relationship and end with some conjectures.

1 2: BASIC NUMBER THEORY RESULTS

The following facts make our exposition of arithmetic self contained. For more a more complete and detailed treatment consult [10] or [13].

Deﬁnition 1. Let a and b be integers, b 6= 0. Then we say b divides a if bq = a for some integer q.

We abbreviate this by writing b|a.

Deﬁnition 2. Let a and b be integers, say a 6= 0. Then the largest positive integer d such that d|a

and d|b is called the greatest common divisor and is denoted gcd(a, b) = d.

Deﬁnition 3. Suppose that a and b are integers such that gcd(a, b) = 1. Then we say that a and b

are coprime or, synonymously, relatively prime.

Theorem 1. Let a, b, and c be integers. Then if b|ac and gcd(a, b) = 1, then b|c.

Proof. Immediate from deﬁnitions.

Theorem 2 (Division Lemma). Given integers a and b > 0, then there exist unique integers q and

r such that

a = bq + r, 0 ≤ r < b, (2.1) where r = 0 if and only if b|a.

2 Proof. Consider the sequence of integers

..., a − 2b, a − b, a, a + b, a + 2b, ... . (2.2)

From this sequence select the smallest non-negative element and denote it r. Hence, r = a − qb

and r = 0 or 0 < r < b, otherwise r was not chosen to be least non-negative element.

To prove uniqueness of q and r, suppose that there is another pair q0, r0 satisfying the same

conditions. Without loss of generality, suppose r0 > r. Therefore, 0 < r0 − r < b and r0 − r =

(q0 − q)b, implying b|(r0 − r). But this is impossible since 0 < r0 − r < b. Therefore, r0 = r and

q0 = q.

Applying Theorem 2 repeatedly, we have Euclid’s Algorithm, a process worthy of study in

its own right. We describe explicitly the algorithm and prove an important property in the next

theorem.

Theorem 3 (Euclid’s Algorithm). Given integers a, b with b > 0, let

a = bq1 + r1, 0 < r1 < b,

b = r1q2 + r2, 0 < r2 < r1,

r1 = r2q3 + r3, 0 < r3 < r2 . .

rk−2 = rk−1qk + rk, 0 < rk < rk−1,

rk−1 = rk + qk+1.

Then gcd(a, b) = rk, the least non-zero remainder in the algorithm described above.

Proof. Induction.

3 Theorem 4 (Bezout’s Lemma). Let a, b with b > 0 be integers. Then there exists integers x, y such that ax + by = gcd(a, b).

Proof. Follows from Theorem 3.

Deﬁnition 4. We call an integer p > 1 prime if there is no divisor d of p such that 1 < d < p.

Theorem 5. Every integer n > 1 can be expressed as a product of primes.

Proof. If n is prime, then we are done. If n is not prime, then it can be factored into n = n1n2

where 1 < n1, n2 < n. If n1 and n2 are prime, we are ﬁnished. Otherwise, repeat process with n1

and n2. This process must terminate because each factor is greater than 1 and less than n.

Theorem 6 (Unique factorization). Let n be positive integer. Suppose that p1, ..., pk are ordered

m1 mk u1 uj and distinct primes, and likewise for p1, ..., pj. Then if n = p1 ··· pk = p1 ··· pj , then k = j

mi ul and pi = pl if and only if i = l.

Proof. Follows from the fact that any two non-zero integers have a greatest common divisor. As- sume non-uniqueness and derive a contradiction.

Theorem 7 (Euclid’s Lemma). If a prime number p divides the integer ab, then p|a or p|b.

Proof. If p does not divide a, then gcd(a, p) = 1. Hence, b = b gcd(a, p) = gcd(ab, pb). Since

p|ab by assumption, and we see p|b.

Theorem 8 (Euclid). There are inﬁnitely many prime numbers.

Proof. Assume there are only ﬁnitely many prime numbers p1, ..., pk. Consider the number

N = p1p2 ··· pk + 1,

a number clearly larger than 1. Hence, there is a prime pi for some 1 ≤ i ≤ k such that pi|N.

Therefore, pi|p1 ··· pk, so pi|N. Hence, pi|(N − p1 ··· pk) = 1, a contradiction.

4 3: IRRATIONALITY

We start with the deﬁnition and some concrete observations.

Deﬁnition 5. A number α is rational if α = p/q for p, q ∈ Z, q 6= 0. Otherwise, α is irrational.

It is readily observed that rational numbers are dense in R. In other words, for any chosen > 0 and real number α we can ﬁnd some rational number p/q such that |α − p/q| < . Informally, it is just there exists a rational number arbitrarily close to any real number we choose. If we are interested in properties of the numbers we are approximating with, density alone does not tell us much. But we could ask this same question, now applying restrictions to . For example, suppose we make the value of dependent on p/q. This changes the problem to a study of the relative closeness of p/q to α. Only a minor change in the problem can lead to some surprising results.

Indeed, making = 1/q2 opens up some deep mathematics that gives information about the local behavior of numbers in the continuum.

An early theorem was stated and proven by Dirichlet in 1842 (see [20], p.34). We replicate it after ﬁrst introducing some useful functions. We introduce the ﬂoor function and its companion, which we will use in the proof of the following theorem.

Deﬁnition 6. For any real number α, we denote bαc = k to be the unique integer k such that k ≤ α < k + 1. Furthermore, we let {α} = α − bαc. We call these the ﬂoor and fractional part of

α, respectively.

5 Theorem 9 (Dirichlet). Let α be any real number and N be any positive integer. Then there exists

a rational number p/q such that 0 < q ≤ N and

p 1 α − ≤ . (3.1) q (N + 1)q

j j+1 Proof. First, we write [0, 1) as the union of N + 1 disjoint sub-intervals: writing Ij = [ N+1 , N+1 ), then

N [ [0, 1) = Ij. j=0

Deﬁne n o S = {α}, {2α}, ..., {Nα} ,

then S is a set of N real numbers in [0, 1) and thus each falls into some Ij. Considering extremal

intervals ﬁrst, suppose for some q ∈ Z that 1 ≤ q ≤ N where {qα} ∈ I0. Then

1 |qα − p| ≤ , for p = bqαc. (3.2) N + 1

Similarly, suppose for some q ∈ Z that {qα} ∈ IN . Then

1 |qα − p| ≤ , for p = bqαc + 1. (3.3) N + 1

Dividing either inequality by q, we have a rational number p/q satisfying (3.1).

If none of the elements in S are contained in I0 or IN , then by the pigeon hole principle, there

exist k, l ∈ Z satisfying 1 ≤ k < l ≤ N such that {kα}, {lα} ∈ Ij, j 6= 0 and j 6= N. Let

6 q = l − k, and p = [lα] − [kα], then

|qα − p| = |(l − k)α − ([lα] − [kα])| (3.4)

= |{lα} − {kα}| (3.5) 1 < (3.6) N + 1 and implying (3.1).

Dirichlet’s argument gave the ﬁrst widely recognized use of the pigeon hole principle [8], p.

2. Since this result is fundamental to the direction we will take, we should try to offer some

interpretation of it. Suppose we have already chosen our α ∈ R. We may freely choose any bound N ∈ N that we like. Dirichlet’s theorem answers the question, “given my choice N, what is the difference between α and at least one rational number p/q ?” The answer is, we can always guarantee |α − p/q| < 1/(N + 1). But in some cases, with the difference we can get as near as

|α − p/q| < 1/(N + 1)2. This invites further investigation just to clarify for which numbers we can guarantee the better estimates. Next, we give two corollaries of Dirichlet’s theorem. Proofs follow the discussion given in [1],pp. 3 - 7.

Corollary 1. Suppose α is irrational. Then there exist an inﬁnitely many rational numbers p/q

satisfying α − p/q < 1/q2.

Proof. We assume that α is an irrational number. Suppose that there are only k rational numbers

p1/q1, ..., pk/qk satisfying p 1 i α − < 2 , (3.7) qi q

7 We deﬁne µ = min{|α − pi/qi| : i = 1, ..., k}. Since α is not rational, then |α − pi/qi| > 0.

Therefore, we can ﬁnd a sufﬁciently large integer n ∈ N, such that

1 0 < < µ. n + 1

By Dirichlet’s theorem, there exists an N, such that

r 1 α − ≤ (3.8) s (N + 1)s

for some rational number r/s. By assumption, then,

r 1 α − < s s2

and hence r/s = pi/qi for sure; but by our choice of N, we have

r α − < µ. s

Therefore, by construction, r/s 6= pi/qi, and so we have reached an absurdity by assuming there are only a ﬁnite number of rational elements that satisfy (3.8). This completes the proof.

Fatou proved a converse to this theorem. For details see [8], p. 11.

Corollary 2. For a rational number α such that α 6= p/q, the inequality α − p/q < c/q2 is

satisﬁed by only ﬁnitely many p/q for any c > 0.

Proof. First suppose that α = r 6= p and that r − p < c . We can see that s q s q q2

1 r p |rq − sp| ≤ − = sq s q sq

8 p c by just considering that |rq − sp| ∈ N. Since by assumption we have α − q < q2 we can write

1 c sq < q2 , but this is true if and only if q < sc.

By Dirichlet’s theorem, 1 ≤ q ≤ N for our chosen bound N. Thus there are only ﬁnitely many q,

and once ﬁxed, only ﬁnitely many p for which this is true. The corollary follows.

The reader might want to consider the following questions. Given some irrational α, what is

the rate of growth rate of the elements p/q satisfying

p 1 α − < ? (3.9) q q2

Also, with respect to some rational α, what is the number of rational numbers p/q such that

p 1 α − < ? (3.10) q q2

As it stands, Dirichlet’s theorem gives no obvious answer to such questions. The latter question is

related to Euclid’s algorithm. In the section on continued fractions we will put forth some partial

answers.

There are many ways to prove that a number is irrational. Here we will give one of the earliest

irrationality arguments (see [13], p.47). As the story goes, the prover was thrown over the side of

a boat to drown by Pythagoras and members of his cult because it contradicted their belief in a

rational universe; for discussion and historical references consult [12], p. 1.

√ Theorem 10. The number 2 is irrational.

√ Proof. Suppose that 2 = p/q where p, q ∈ Z, q 6= 0, and (p, q) = 1. Then p2 = 2q2, which implies that p is even. Since p = 2r, p2 = 4r2 = 2q2. Hence, q2 = 2r2. So q too is even. But then

9 √ p and q share a common factor namely 2, so contradicting the lowest-terms hypothesis. Hence, 2

is irrational.

A notable feature of this argument is that it relies on a contradiction where divisibility is at

the heart of the matter. But one could ﬁnd reason to be dissatisﬁed if they suspect the property

of being irrational is independent of whether we are dividing a number by a reduced fraction √ or not. So utilizing Dirichlet’s theorem, we will show that the golden ratio φ = (1 + 5)/2

is irrational with an alternative argument. Note that φ and −1/φ are the distinct roots of the

polynomial P (z) = z2 − z − 1. We shall see in another section that this method of proof can be √ generalized. Indeed, a proof that 2 is irrational can be cast exactly the same way.

The next important sequence will have many appearances in this article.

Deﬁnition 7. We deﬁne the Fibonacci sequence as F0 = 0,F1 = 1 and Fn = Fn−1 + Fn−2 for all n ≥ 2.

Theorem 11. The golden ratio φ is irrational.

We will rely on two very simple properties of the Fibonacci numbers, each given in a lemma.

th Lemma 1. If Fn is the n Fibonacci number, then for all n ≥ 0

2 n Fn+1Fn−1 − Fn = (−1) . (3.11)

10 1 2 n Proof. We use induction. First, 1·0−1 = (−1) . Now assume Fn+1Fn−1 −Fn = (−1) . Consider

2 2 Fn+2Fn − Fn+1 = (Fn + Fn+1)Fn − Fn+1 (3.12)

2 = Fn + Fn+1(Fn − Fn+1) (3.13)

2 = − (Fn+1 − Fn)Fn+1 − Fn (3.14)

2 = − Fn−1Fn+1 − Fn (3.15)

= −(−1)n = (−1)n+1. (3.16)

Lemma 2. If Fm/Fm−1 < φ < Fn/Fn−1, then m is even and n is odd.

Proof. Recall that φ is the larger of two roots of the polynomial P (z) = z2 − z − 1. Suppose

2 φ < Fm/Fm−1. Then by substituting Fm/Fm−1 in x − x − 1 we know

F 2 F m − m − 1 > 0. Fm−1 Fm−1

2 Next we multiply by Fm−1 to get

2 2 0 < Fm − Fm−1Fm − Fm−1

2 = Fm − Fm−1(Fm + Fm−1)

2 = Fm − Fm−1Fm+1

= −(−1)m

= (−1)m+1.

Hence, it is impossible that φ < Fm/Fm−1 when m is even. The proof for n odd is the same.

11 Proof of Theorem 11. By Lemma 1 we have

2 n Fn+1Fn−1 − Fn = (−1) . (3.17)

Multiplying both sides by 1/FnFn−1 we get

Fn+1 Fn 1 − = . (3.18) Fn Fn−1 FnFn−1

Specializing Lemma 2, we have, say Fn/Fn−1 < φ < Fn+1/Fn. Manipulating we get

Fn Fn+1 Fn 1 1 φ − < − = < 2 . (3.19) Fn−1 Fn Fn−1 FnFn−1 Fn

By Corollary 2, we have shown φ is irrational.

So far we have seen two proofs that a number is irrational. Questions of irrationality are

still very much alive. In the forthcoming, we will develop some powerful tools to investigate

the character of particular numbers, but all these tools are essentially classical. We note that many

numbers have resisted classical approaches. As one would expect modern tactics have solved some

problems that were before out of reach; however, some numbers have resisted all approaches. For

example, it is still unknown whether the Euler-Mascheroni constant

n ! X 1 γ = lim − ln(n) , (3.20) n→∞ k k=1

is irrational ([18], p.45). There are no apparent patterns in the decimal expansion of γ, which is approximately equal to

0.57721566490153286060651209008240243104215933593992.

12 Other fundamental constants have also resisted interpretation. For instance,

∞ X 1 ζ(2k + 1) = (3.21) n2k+1 n=1

for k ∈ N represents a class of numbers that have deﬁed approaches to classiﬁcation. For further details see recent papers [19] and [27]. It was only recently in 1978 that Apery´ showed ζ(3) is irrational using Dirichlet’s condition for irrationality. An outline is given in [24], p. 52. In this paper we will discuss more irrationality proofs for some interesting numbers.

13 4: TRANSCENDENCE

We begin by reviewing some deﬁnitions and theorems.

d d−1 Deﬁnition 8. Suppose d is a natural number. Let p(z) = adz + ad−1z + ... + a0 with ai ∈ Z be an irreducible polynomial. If there exists an α ∈ R such that p(α) = 0, then we say α is algebraic; otherwise α is called transcendental.

We state an important theorem ﬁrst rigorously proven by Argand.

d d−1 Theorem 12 (Fundamental Theorem of Arithmetic). If p(z) = adz + ad−1z + ... + a0 ∈ C[z] is a polynomial of degree d, then f(z) has at most d roots in C.

We shall also use a result from elementary calculus.

Theorem 13 (Mean Value Theorem). If f is continuous on the closed interval [a, b] where a < b and differentiable on (a, b), then there exists some c ∈ (a, b) such that f 0(c) = (f(b) − f(a))/(b − a).

Now, with those in place, we proceed. If a number is rational, then it is evident we can make

it the root of a polynomial with integer coefﬁcients. So rational numbers are clearly algebraic.

However, this question is not easy if given an arbitrary irrational number. By Dirichlet’s theorem,

we know we can ﬁnd inﬁnitely many “good” rational approximations to α only if α is irrational.

In fact, irrationality and “transcendentality” are two sides of the same coin. It turns out that the

question of whether α is transcendental is just a question about how well we can approximate it

14 by inﬁnitely many rational numbers. Transcendental numbers allow advantages of rational ap-

proximation over algebraic numbers. The connection of zeros of polynomial and irrationality is

surprising and for this result we owe Liouville [1], p. 5. We will now work towards a proof of a

precise statement of this.

Theorem 14 (Liouville). Let α be irrational and algebraic for an irreducible polynomial of degree d. Then there exists a constant C(α) dependent exclusively on α such that for all p/q ∈ Q,

p C(α) α − > . (4.1) q qd

Proof. Suppose α is algebraic deg(f) = d. Using the fundamental theorem of arithmetic, if

α, α1, ..., αd−1 are the roots of f, let β = αj where 0 < |α − αj| ≤ |α − αi|, i = 1, ..., d − 1. We let δ = |α − β| > 0.

δ/2 If p/q is arbitrary, suppose that α−p/q ≥ δ. Consider the quantity δ/2. Then |α−p/q| > qd . Therefore, we may allow |α − β|/2 = C(α).

p Next, suppose that α − q < δ. Then we know by our choice of β that p/q is not a root of f(x). Actually, it is easily seen that, when combining under a single denominator, we have

p a pd + a pd−1q + ... + a p0qd 1 f = d d−1 0 ≥ . q qd qd

Consider the interval (α − δ, α + δ). Appealing to the mean value theorem, we know that there exists some γ ∈ (α, p/q) such that

f(p/q) − f(α) f(p/q) = = f 0(γ) 6= 0, (4.2) (p/q) − α (p/q) − α

15 where γ ∈ (α, p/q). Using the assumption that α∈ / Q, we see that

p p 1 f = α − · f 0(γ) ≥ q q qd

Thus we see that p 1 α − ≥ . q |f 0(γ)|qd

Now select K such that K > |f 0(x)| for all x ∈ (α − δ, α + δ). Then letting C(α) = c = 1/K we have the strict inequality needed for the theorem i.e.

p c α − > . (4.3) q qd

Thus from all cases we have show how to construct the C(α) of the theorem and hence ﬁnished.

Using theorem 14, as a corollary Liouville constructed the ﬁrst transcendental number in 1844

[1], p. 6. Transcendental numbers had been conjectured to exist since Leibniz or earlier, but this

was the ﬁrst explicit transcendental number. Liouville’s idea was based on the implication, given

some real number α and any choice of d, if we can ﬁnd inﬁnitely many rational numbers p/q such

that p 1 α − < , (4.4) q qd then α must be transcendental.

Corollary 3. The number ∞ X 1 (4.5) 10k! k=1 is transcendental.

16 P∞ k! Pn k! Proof. Suppose that L = k=1 1/10 and Ln = k=1 1/10 . Assume L is algebraic of degree d. First we must demonstrate that d > 1. We know a number is rational when its deci-

mal expansion terminates or repeats. Observe L1 = 0.1, L2 = 0.11,L3 = 0.110001,L4 =

0.110001000000000000000001... . Clearly, the decimal expansion of L is not ﬁnite. An easy in-

ductive argument on Ln shows that the decimal expansion of L is not repeating, hence we know

that d > 1 or that L is not rational.

Next, we know that if L is algebraic, then it has degree greater than or equal to 2. Hence, by

Liouville’s theorem there exists some C(L) such that

p C(L) L − > (4.6) q qd

n! for all p/q ∈ Q. Now we construct a sequence of rational approximations by deﬁning pn = 10 Ln

n! and qn = 10 , giving p ∞ 1 n X L − = k! . qn 10 k=n+1 Consider, by the size of denominators, how

∞ ∞ X 1 X 1 < (4.7) 10k! 10k k=n+1 k=(n+1)! ∞ 1 X 1 k = (4.8) 10(n+1)! 10 k=0 1 1 = . (4.9) 10(n+1)! 1 − 1/10

Hence, we have p 10 1 n L − < (n+1)! . qn 9 10

17 Using (4.6) and the bound we have

C(L) 10 1 < , 10n!d 9 10(n+1)!

an inequality that can be rewritten as

10(n+1)! 10 1 < . (4.10) 10n!d 9 C(L)

But there is an N such that for n > N statement (4.10) is absurd. Hence, L is transcendental.

Though L is transcendental, L is artiﬁcial in the sense it was constructed to meet requirements

of theorem (14). It is not an uninteresting number; however, a more interesting case would be

to determine that a familiar constant is transcendental. We will see eventually that a very natural

number, namely e, is transcendental.

Finally, we only mention an important result due to Roth that built on decades of work, begun

by Liouville and continued by mathematicians such as Dyson, Siegel and Thue [24], p. 38. Roth’s

theorem has been generalized to higher dimensions by Schmidt (see [20]). We state it now as given

in [6]:

Theorem 15 (Roth). Let α be an algebraic number and > 0 be arbitrary. There are only ﬁnitely many p/q such that p 1 α − < . (4.11) q q2+

Proof. See [20], pp. 39-57, and [24], pp. 162-173.

Recall that for Liouville’s number L, we saw that, given any natural number n, there exists a

rational number p/q such that p 1 L − < . (4.12) q qn 18 Hence it easily follows from Roth’s theorem that L is transcendental. But it also prompts additional questions. In other words, L seems like perhaps the “best” case since the exponent is unbounded.

To give a quantity for this notion, we introduce the following deﬁnition.

Deﬁnition 9. Let Ω(q) be a positive function of q ∈ Z that approaches 0 as q approaches inﬁnity. Let > 0 be arbitrary. Then we say α is approximable to order Ω(q) by rational numbers if there exist a positive constant C(α) = C dependent only on α such that

p α − < (C − )Ω(q) (4.13) q

has only ﬁnitely many solutions p/q ∈ Q and

p α − < (C + )Ω(q) (4.14) q

has inﬁnitely many solutions p/q ∈ Q.

Now, ﬂeshing out the deﬁnition, by Roth’s theorem every irrational number must have an order

of approximation that is Ω(q) = 1/q2. Also an immediate corollary of Roth’s theorem is that the

order of approximation to L is inﬁnite, since any functional bound is insufﬁcient. Any number

with an inﬁnite order of approximation is called a Liouville number.

It turns out that using measure theoretic methods one can show Liouville numbers are extremely

rare [5], p. 352. We will show in a later section the order of approximation of e. The order of

approximation to π is still unknown [5], p. 371. In general, establishing the order of approximation

Ω(q) for a particular number α is very difﬁcult.

19 5: ELEMENTARY CONTINUED FRACTIONS

Deﬁnition 10. Let α be a real number. Let I = {0, 1, ..., n} or I = N. Then by a simple continued fraction we mean a fraction of the form

1 α = a0 + , (5.1) 1 a1 + a2 + ...

where ai ∈ N for all i ∈ I.

∞ n That we can do this follows from Theorem 2. When is needed we write (ai)i=0 or (ai)i=0 to

make the index set I explicit. Also, nothing prohibits us from taking (ai)i∈I to be a sequence of

real or complex numbers, but then questions of convergence must be addressed [21], p. 335; more

general cases such as this, however, are no longer called simple continued fractions. Simple con-

tinued fractions always converge. In this paper we will only be concerned with simple continued

fractions and often we will drop “simple” from the deﬁnition and just write “continued fraction.”

th Deﬁnition 11. An element ai of the sequence (ai)i∈I is called the i partial quotient of the contin-

ued fraction.

Given the deﬁnition 10, we can denote the continued fraction of a number α as a list of its

partial quotients

α = [a0; a1, ..., an], (5.2)

20 or, where the sequence (ai) is inﬁnite, we may write

α = lim [a0; a1, ..., an]. (5.3) n→∞

Deﬁnition 12. If α = [a0; a1, a2, ...], then we say sequence (ai) is associated with α.

Numerous notations exist for representing fractions as some notation can be an impediment in

calculations and an asset in others; our notation is orthodox. Fortunately, there is an easy recursive

deﬁnition for the rational number p/q = [a0; a1, ..., an]. We agree to write

pn = [a0; a1, ..., an], (5.4) qn where

pn = anpn−1 + pn−2 and qn = anqn−1 + qn−2 (5.5)

with initial conditions p−1 = 1, q−1 = 0, p0 = a0 and q0 = 1.

Deﬁnition 13. Given (ai)i∈I associated with α, then for any non-negative integer we call pk/qk =

th [a0; a1, ..., ak] the k convergent to α.

In contrast to deﬁnition 13, we also deﬁne complete quotients, or the “tails” of the continued

fraction.

∞ ∞ Deﬁnition 14. Let (ai)i=1 be associated with α. Then for any k we call (ai)i=k a complete quotient

of α. We often denote this as [ak, ak+1, ...] = αk.

There are two common ways to ﬁnd the convergents of a continued fraction. The ﬁrst is re-

cursion that deﬁnes the class of continuant polynomials. In [15], p. 15 they are called the Euler-

Mindingen formulas. We present it in a deﬁnition as they are presented in [12], p. 162.

21 f2 = x1x2 + 1 f3 = x1x2x3 + x1 + x3 f4 = x1x2x3x4 + x1x2 + x1x4 + x3x4 + 1

Table 5.1: Continuant polynomials

Deﬁnition 15. Let f0 = 1 and f1(x1) = x1. Then a continuant polynomial is deﬁned by the

recursion

fk(x1, x2, ..., xk) = xkfk−1(x1, x2, ..., xk−1) + fk−2(x1, x2, ..., xk−2). (5.6)

We give the ﬁrst few for convenience.

Theorem 16. Let (ai) be associated with α and pn/qn a convergent to α. Then

f (a , ..., a ) p n 0 n = n . (5.7) fn−1(a0, ..., an−1) qn

Proof. See theorem 1.8 in [15], pp. 15 - 16.

The most popular method of ﬁnding convergents is due to Hurwitz [6], p. 24, which we state

in the following theorem.

Theorem 17 (Hurwitz). Let (ai) be associated with α. If pn/qn is a convergent to α, then

        a 1 a 1 a 1 p p  1   2   n   n n−1     ...   =   . (5.8) 1 0 1 0 1 0 qn qn−1

Proof. From (5.5) and the initial conditions, (5.8) easily follows by induction.

22 Here we remark on notation. For the remainder of the thesis we will have a strong interest in

convergents. Therefore, for readability we abbreviate “pn/qn is a convergent to the real number α”

by writing “pn/qn ,→ α”.

The properties of pn/qn form a rich source. We will prove a few of these properties, but we are under the shadow of a vast literature; however, there are also many simply stated open questions regarding such series. For example, recall the Fibonacci sequence that we introduced in deﬁnition

7. These numbers result as a specialization of the continuant polynomials. It is still unknown whether inﬁnitely many of the Fibonacci numbers are prime.

Next, let α be any irrational number. We demonstrate an algorithm that will construct a simple continued fraction expansion of α. This can be thought of just as an extension of Euclid’s algorithm that fails to terminate when α is not rational. If bαc and {α} are the ﬂoor and fractional parts of α, respectively, the continued fraction expansion of α is constructed iteratively as shown:

1 α = bαc + . (5.9) 1 b{α}−1c + b{{α}−1}−1c + ...

A number α is irrational if and only if can represented as an inﬁnite simple continued fraction.

This is a straightforward observation since the fractional part of a real number is irrational if and

only if the real number itself is irrational. Hence, we can give a second proof of the irrationality of √ φ = (1 + 5)/2 by showing a sequence associated with φ is inﬁnite.

Proposition 1. The sequence [1, 1, 1, ...] is associated with φ.

Proof. Letting a0 = p0 = bφc = 1, we have pn = Fn and qn = Fn−1 by the equation (5.5) and Deﬁnition 7. Since pn = Fn = Fn−1 + Fn−2, then pn < 2Fn−1. Therefore, bFn/Fn−1c ≤

23 2Fn−1/Fn−1 = 2. Thus, bFn/Fn−1c = 1. By continued fraction algorithm (5.9), the conclusion

follows.

We note that, similar to decimal expansions of numbers, partial convergents can exhibit periods.

In the case above we may write φ = [1].

We next prove some properties of convergents.

Proposition 2. Suppose (ai) is associated with α, and {pi/qi}i∈I are convergents to α. Then

n 1. qnpn−1 − pnqn−1 = (−1) ;

n−1 2. qnpn−2 − pnqn−2 = an(−1) .

Proof. For the ﬁrst statement, this follows from Theorem 17. Observe

  a 1  n  det   = −1. 1 0

Therefore,   p p  n n−1 n det   = (−1) qn qn−1 and the statement is proved.

Note that the ﬁrst statement gives

p p (−1)n p p (−1)n−1 n−1 − n = and n−2 − n−1 = . (5.10) qn−1 qn qn−1qn qn−2 qn−1 qn−2qn−1

24 Thus, by this and (5.5) we have

p p 1 1 n−2 − n = (−1)n−1 − qn−2 qn qn−2qn−1 qn−1qn

q − q = (−1)n−1 n n−2 qn−2qn−1qn

a = (−1)n−1 n qn−2qn and the second statement follows.

The ﬁrst statement of Proposition 2 yields the following corollary.

Corollary 4. Suppose pn/qn = [a0; a1, ..., an]. Then gcd(pn, qn) = 1.

Proposition 3. Let an irrational number α be given. Suppose pm/qm, pn/qn ,→ α. Then if pm/qm < α < pn/qn, then m is even and n is odd.

Proof. Observe that Proposition 2 implies

p p (−1)n n−1 − n = . (5.11) qn−1 qn qn−1qn

2 Therefore, if |α − pn/qn| < 1/qn, we know that

p p n−1 < α < n (5.12) qn−1 qn

for n even or n odd. Recalling limn→∞[a0, a1, ..., an] = α, observe that

n−1 pn 1 1 (−1) = a0 + − + ... + . (5.13) qn q0q1 q1q2 qn−1qn 25 Thus, p 1 1 α − n = (−1)n − + ... , (5.14) qn qnqn+1 qn+1qn+2

and it follows that p 1 α − n < . (5.15) 2 qn qn

Now, by an easy induction on n, the claim follows.

Propositions 2 and 3 together imply the following important result.

Theorem 18. The even and odd convergents form monotonic sequences, from below and above respectively.

Proposition 4. Let (ai) be associated with α and suppose that pn/qn ,→ α. Then

1 p 1 < α − n < . (5.16) 2 2 (an+1 + 2)qn qn an+1qn

Proof. From Proposition 2, ﬁrst statement, and Proposition 3 we have

p p p 1 1 α − n < n − n−1 = ≤ , (5.17) 2 qn qn qn−1 qnqn+1 an+1qn

where the last inequality follows from (5.5).

For the lower bound, from Corollary 18 we get

p p a n − n+2 = (−1)n+1 n+2 . (5.18) qn qn+2 qnqn+2

Thus we have p p p a 1 α − n ≥ n − n+2 = n+2 > , (5.19) 2 qn qn qn+2 qnqn+2 (an+1 + 2)qn

again with use of (5.5).

26 We pause to say something about Theorem 4. The reader is invited to compare this result to

Dirichlet’s theorem. Continued fraction convergents give a constructive realization of Dirichlet’s

theorem. Furthermore, continued fraction convergents are the best we can do to approximate an

irrational number in the following sense:

Proposition 5. Let α be irrational and q an arbitrary natural number. Let n be the largest such that pn/qn ,→ α satisﬁes qn ≤ q. Then

p s α − n ≤ α − . (5.20) qn t

This following theorem is due to Vahlen (see[7], p., [8], p., and [15], p.31).

Theorem 19 (Vahlen). Let α be irrational. Then if pn and pn+1 are convergents to α, then qn qn+1

p 1 α − n+i < (5.21) 2 qn+i 2qn+i

for i = 0 or i = 1.

Proof. Suppose that 19 does not hold for i = 0 or i = 1. Then

1 1 p p p p 1 + ≤ α − n + α − n+1 = n − n+1 = , (5.22) 2 2 2qn 2qn+1 qn qn+1 qn qn+1 qnqn+1

which contradicts the fact that qn+1 > qn. √ Borel showed we can replace 2 with 5 in Vahlen’s theorem. There is an important theorem,

due to Legendre that is a partial converse of theorem 19 [8], p. and [15], p. 31.

Theorem 20 (Legendre). Suppose that for some irrational number α we have

p 1 α − < . (5.23) q 2q2 27 Then p/q = pn/qn for some natural number n.

In a similar vein, Hurwitz extended Vahlen’s theorem; see [8], p. 20.

Theorem 21 (Hurwitz). Let α be irrational. Then if pn , pn+1 and pn+2 are convergents to α, then qn qn+1 qn+2

p 1 α − n+i < √ qn+i 5q2

for i = 0, i = 1, or i = 2.

Hurwitz’ result is the best we can do for an arbitrary α.

Theorem 4 gives some justiﬁcation for Deﬁnition 16; this turns out to be a very natural condition.

Deﬁnition 16. The irrational number α is badly approximable if all quotients are absolutely bounded.

We have seen in Theorem 1 that φ has period one. Hence, the partial quotients are bounded and thus φ is badly approximable. It is known that e, the base of the natural logarithm, has the associated sequence [2, 1, 2, 1, 1, 4, 1, 1, 6, ...]. Thus, by Deﬁnition 16 the number e is not badly

approximable. In fact, the number e is not a Liouville number either. This can be seen by the

following result of Davis (c.f. [6], pp. 49 - 51).

Theorem 22 (Davis). Let e be the base of the natural logarithm. Then for any > 0, there exist

only ﬁnitely many p/q satisfying

p 1 ln ln q e − < − ; (5.24) q 2 q2 ln q

28 conversely, there exist inﬁnitely many p/q satisfying

p 1 ln ln q e − < + ; (5.25) q 2 q2 ln q

An immediate corollary of Roth’s theorem is that e is transcendental. The ﬁrst proof that e is transcendental was given by Hermite in 1873 [5], p. 348. Hermite’s proof is considerably more difﬁcult without the advantage of Roth’s theorem.

29 6: LAMBERT SERIES, ϑ-FUNCTIONS, AND LANDAU’S RECIPROCAL FIBONACCI SUM

For a given irrational number α, let pn/qn ,→ α. In this chapter we shall discuss the series

∞ X 1 . (6.1) q n=0 n

The ﬁrst example is due to Landau [16], so we call any series of the form (6.1) a Landau sum. If

th Fn is the n Fibonacci number, then

∞ X 1 ψ = = 3.359885666243177... . (6.2) F n=1 n

We introduce a deﬁnition based on a geometric interpretation of the numbers 1/q. For example,

2 letting α be irrational, and consider the line αx ∈ R . Then, if P = (pn, qn) and Q = (pn+1, qn+1) are points such that p 1 α − n+i < 2 qn+i qn+i

for i = 0 and i = 1, then the line passing through P and Q will intersect the y-axis at ±1/qn+1

th Deﬁnition 17. Let pn/qn ,→ α. Then we call 1/qn the n projection of α if qn 6= 0.

30 This is sometimes called Landau’s constant or the reciprocal Fibonacci constant [26]. Proof of

the representation of ψ as a ϑ-function will follow from two theorems ﬁrst shown by Landau. It is

the aim of this section to prove Landau’s theorems.

Landau’s result is seemingly not widely known; however, is extremely interesting. In general,

it is very hard to say anything non-trivial about the sum 6.1. For example, take the well known

constant e as an example. We know that

e = [2; 1, 2, 1, 1, 4, 1, 1, 6, ...] (6.3) or, equivalently,

e − 1 = [1; 1, 2, 1, 1, 4, 1, 1, 6, ...], (6.4) which is notable for its very nice pattern. The sum 6.1 for e, however, does not appear to have a nice representation such as what Landau showed for ψ; nonetheless, reciprocal sums for numbers such as e do not appear to be far out of reach. But ﬁnding such a sum for a real number with unpredictable partial quotients, such as π for instance, seems very hard. Before we start working with Landau’s sums, we want to state and prove a result that is classical and widely known, but nonetheless fundamental to everything we do in the sequel. Also, it will serve as a special model of a more general constructions later. It is called Binet’s formula [9], p. 237, but evidently it was known earlier by both D’Alembert and Bernoulli [14].

√ √ Theorem 23 (Binet). Let φ = (1 + 5)/2 and κ(φ) = (1 − 5)/2. Then the nth Fibonacci

number Fn is given by φn − κ(φ)n F = . (6.5) n φ − κ(φ)

31 Proof. The Fibonacci numbers are deﬁned by Fn = Fn−1 + Fn−2 where F0 = 0 and F1 = 1. We apply the technique of constructing a generating function to give a proof. We have

X n X n Fnx = (Fn−1 + Fn−2)x (6.6)

X n−1 2 X n−2 = x Fn−1x + x Fn−2x . (6.7)

n 2 Since our goal is to isolate a single term Fnx we multiply by a ‘normalizing’ factor F2x . We have

X n 2 X n−1 2 X n−2 2 Fnx (F2x ) = x Fn−1x + x Fn−2x (F2x ), (6.8)

But this we can account for in the indices to produce a like term. That is, we get

∞ ∞ ∞ X n X n 2 X n Fnx = x Fnx + x Fnx , (6.9) 2 1 0

or just utilizing the initial values F0 and F1 we have

X n X n 2 X n Fnx − F1x − F0 = x Fnx − F0 + x Fnx (6.10)

that we rewrite again, now satisfactorily, as

∞ X x F xn = . (6.11) n 1 − x − x2 n=0

Equality (6.11) is what is useful to ﬁnd the explicit formula we desire. For notice that

x x = (6.12) 1 − x − x2 (1 − φx)(1 + κ(φ)x)

32 for φ yet determined. We can write φ = (a + b); now multiplying (a + b)κ(a + b) = a2 − b2 and taking the difference −(a + b) − κ(a + b) = −2a, we may write

x x = , (6.13) (1 − φx)(1 + κ(φ)x) 1 − 2ax + (a2 − b2)x2

1 giving us something solvable. See that a = 2 is determined and thus we are left to solve for b. We √ √ solve 1/4 − b2 = −1 and get b = 5/2. Therefore φ = (1 + 5)/2.

We are close to what we want. We have by partial fraction decomposition of the right hand side

of (6.12),

x A B = + . (6.14) (1 − φx)(1 + κ(φ)x) (1 − φx) (1 + κ(φ)x)

Now solving (1 − κ(φ))xA + (1 − φx)B = x in steps:

√ √ A(1 − 5) − B(1 + 5) (1 − κ(φ)x)A + (1 − φx)B = A + B − x 2 √ √ A(1 − 5) − A(1 + 5) = A − A − x 2 √ √ −A 5 − A 5 = − x. 2

33 √ √ And thus by observation A = 1/ 5 and therefore B = −1/ 5. Hence, using what we know

about the sum of two geometric series, we can write

X n 1 1 1 1 Fnx = √ − √ 5 (1 − φx) 5 (1 − κ(φ)x) 1 X 1 X = √ φnxn − √ κ(φ)nxn 5 5 X φn − κ(φ)n = √ xn 5 X φn − κ(φ)n = xn. φ − κ(φ)

n n Therefore, we have shown that Fn = (φ − κ(φ) )/(φ − κ(φ)). This completes the proof.

We make some observations to motivate the upcoming. Letting 1 > |b| and given q0, we deﬁne qn = qn−1b. Then we see

∞ X 1 1 1 1 1 = + + + + ... q q q b q b2 q b3 n=0 n 0 0 0 0

or just a familiar geometric series recursively deﬁned. Now, suppose that q0, q1 are given and

qn+1 = qnb + qn−1c, we have a sum

∞ X 1 1 1 1 1 = + + + + q q q q b + q c q b2 + q bc + q c n=0 n 0 1 0 1 0 1 1 1 3 2 2 + ... . q0b + q1b + q1bc + q0bc + q1c

These two recursively deﬁned series are very similar; indeed the geometric series is a special case

of the latter. We will see that ϑ-functions (to be deﬁned below) provide the correct analogue of a

geometric series where the denominators are recursive secondary polynomials of degree n − 2 for

n > 3.

34 Deﬁnition 18. Letting |q| < 1, we deﬁne ϑ-functions of a real variable to be

∞ n+1 2 P ( 2 ) 1. ϑ2(q) = −∞ q ϑ2(0) = 0;

P∞ n2 2. ϑ3(q) = −∞ q ϑ3(0) = 1.

The variable q in Deﬁnition 18 is used according to convention; it has nothing to do with

previous material.

This next proposition gives an elementary, though interesting fact. It is essentially the workhorse

of the remaining portion of the thesis.

Proposition 6. Suppose that α 6= β are solutions to x2 − bx − c = 0; i.e. they are roots of the

quadratic equation x2 − bx − c. Set

(q − βq )αn − (q − αq )βn q = 1 0 1 0 . (6.15) n α − β

Then we have

1. α + β = b and αβ = c.

2. qn+1 = qnb + qn−1c

Proof. See [5], p. 91.

For what we need, we restrict to the case of b ∈ R and c = ±1. We meet the conditions of theorem 6 by writing qn+1 = 2uqn + νqn−1 where |ν| = 1, u > max{0, −ν} and assume that √ √ α, β ∈ R, then we know −c = ±1, α = u+ u2 + ν, and β = u− u2 + ν. Therefore, supposing

A = (q1 − βq0) and B = (q1 − αq0), we care about sums of the form

∞ X 1 (α − β) . (6.16) Aαn + Bβn n=1 35 We ﬁnd the bridge between our desired sums and ϑ-functions using Lambert series.

Deﬁnition 19. Let β be a real number such that |β| < 1. A series of the form

∞ X βn = L(β) (6.17) 1 − βn n=1

is called a Lambert series.

Lambert series give us the following identities.

Theorem 24. Letting 0 < β < α such that αβ = 1,

P∞ 1 P∞ βn 1 2 1. n=1 αn+βn = n=1 1+β2n = 4 (ϑ3(β) − 1))

P∞ 1 P∞ β2n 1 2 2 2. n=1 α2n+β2n = n=1 1+β4n = 4 (ϑ3(β ) − 1))

P∞ 1 P∞ β2n+1 1 2 2 3. n=1 α2n+1+β2n+1 = n=1 1+β4n+2 = 4 ϑ2(β )

P∞ 1 2 4. n=1 αn−βn = L(β) − L(β )

P∞ 1 2 4 5. n=1 α2n−β2n = L(β ) − L(β )

P∞ 1 2 4 6. n=1 α2n+1−β2n+1 = L(β) − 2L(β ) − L(β )

Proof. In [5], pp. 91 - 94.

th Theorem 25 (Landau). Where Fn is the n Fibonacci number, then

∞ √ 4 X 1 3 − 5 √ = ϑ2 . (6.18) F 2 2 5 n=0 2k+1 √ 2 Proof. First, note that (1 ± 5)/2 are the roots to x − x − 1. So b = c = 1 and then qn = qn−1 + qn−2. So it must be that u = 1/2. Then α and β are as we desire, and α + β = b, αβ = −c

36 √ P 2n+1 2n+1 P as needed. Recall that by Binet’s formula we know 5/(α − β ) = 1/F2n+1 We thus use (6.16) to write

∞ ∞ X 1 √ X 1 (α − β) = 2 u2 + 1 . (6.19) q α2n+1 + β2n+1 n=1 n n=0

√ q 1 We notice that substitution in u gives us 2 22 + 1 = 5. Hence, using ϑ identity from Theorem 24, we get √ 2 √ √ ∞ √ 1 1 5 5 3 − 5 X 1 5 ϑ2 − = ϑ2 = . (6.20) 4 2 2 2 4 2 2 F n=0 2k+1 This completes the proof.

Similarly, we have the following:

Theorem 26 (Landau). Where Fn is as before,

∞ √ √ 1 X 1 3 − 5 7 − 3 5 √ = L − L . (6.21) F2n 2 2 5 k=1

Corollary 5. We evaluate

ψ ≈ 3.359885666243177. (6.22)

Proof. Combining Theorem 25 and Theorem 26, we may compute ψ to arbitrary precision.

√ It has not been explored by the author, but it is curious that 3− 5/2 appearing in the ϑ-function

is an eigenvalue of the matrix

        0 1 3 8           7→   7→   7→   7→ ... (6.23) 1 1 2 5

37 In [2] it is shown that that ψ is irrational, giving a solution to a conjecture of Erdos.¨ Note that

despite the apparent irregularity in the decimal expansion of (6.2), showing that ψ is irrational is

non-trivial. Put another way, it is not clear on the surface that ψ does not have an exorbitantly

large period in the decimal expansion, since the number is only known through ϑ-computations.

Consider for example a class of inﬁnite non-geometric series whose images under N are strictly rational numbers, namely ∞ X a 1 = H (6.24) n2 + a2 2 2a n=a+1 Pm where Hm = k=1 1/k. Why Landau’s sum and similar sums should be different is an interesting question.

38 7: FIBONACCI POLYNOMIALS AND MORE LANDAU SUMS

The aim of this section is to prove a generalization of Landau’s theorem for ψ. It turns out there

is nothing special about the projections of φ. Indeed, for any number of the form α = [k] for k a

natural number, there exists a ϑ-function representation of the sum of its projections. So in other

words, ψ = ψ1 in the list ψ1, ψ2, ψ3, ... . Some of these numbers we compute to small margin of

accuracy in the ﬁnal section. √ √ Recall that (1 + 5)/2 and (1 − 5)/2 are roots to the equation x2 − x − 1. Now observe √ √ that 1 + 2 and 1 − 2 are roots to x2 − 2x − 1. We have proved already that φ = [1] and it is √ well known and easy to verify that 1 + 2 = [2]. The formal resemblance of the characteristic

polynomial with the continued fraction expansion of its principle root is not a coincidence. It turns

out that if α = [k], then it is a root to an equation x2 − kx − 1. Recalling Deﬁnition 16, these numbers are in a sense the worst we could possibly choose to approximate by rational numbers.

But we will see that they also give us a fair bit of information about the projections of any real number α.

With some motivation in place, we begin working towards a generalized result about Landau sums. It is fairly easy exercise to show, by induction, the following result.

39 Proposition 7. For any natural number k,

√ k + k2 + 4 1 = k + . (7.1) 2 1 k + k + ...

Therefore, given any continued fraction of period one, we have a closed form for limn→ pn/qn.

These constant continued fractions turn up in unexpected places. The paper [17] gives one interesting example.

We give two corollaries that relate φ to constant continued fractions. Before we continue, however, we deﬁne another important sequence of numbers that is known to have a close relationship to the Fibonacci sequence.

Deﬁnition 20. Let L0 = 2 and L1 = 1. Then if Ln + Ln−1 = Ln+1 for a natural number n, then

Ln+1 is called a Lucas number.

√ th Corollary 6. Let φ = (1 + 5)/2 be the golden ratio. Let Ln be the n Lucas number i.e.

Ln = Ln−1 + Ln−2 for L0 = 2,L1 = 1. Then for all positive integers k

2k+1 φ = [L2k+1]. (7.2)

Proof. With Theorem 23 and

√ √ 1 + 5n 1 − 5n L = + (7.3) n 2 2 from [25], p. 52, we combine to get

√ 1 + 5n 1 √ = L + F 5 . (7.4) 2 2 n n

40 We want to show √ 1 + 52k+1 1 q = L + L2 + 4 . (7.5) 2 2 2k+1 2k+1

2 n From [25], p. 177, we use Ln = L2n + (−1) 2, substituting under the radical. This with identity

2 n 5Fn = L2n − 2(−1) , again from [25], p.177, we have

q √ 1 2 1 L2k+1 + 5F2k+1 = L2k+1 + F2k+1 5 (7.6) 2 2 √ 1 + 52k+1 = . (7.7) 2

This is what we wanted.

We remark that the identity in Corollary 6 can be generalized for other powers of numbers [k], with the appropriate linear recurrence dependent on [k].

To continue to the main result, we will use Fibonacci Polynomials. Fibonacci polynomials are

special functions that have application in combinatorics and number theory. These are deﬁned as

the polynomial analogue to Fibonacci numbers. Two good sources on Fibonacci polynomials are

[4] and [18].

th Deﬁnition 21. Let F0(k) = 0 and F1(k) = 1. The n Fibonacci polynomial is given by

Fn(k) = xFn−1(k) + Fn−2(k). (7.8)

We use variable k in anticipation of the upcoming discussion, but the domain of Fn could be anything.

Though they are very easy to compute, we list the next eight for convenient reference. If we take x = 1, then we recover the recursion an = an−1 + an−1 where a0 = 0 and a1 = 1. Hence,

41 5 3 F2(k) = k F6(k) = k + 4k + 3k 2 6 4 2 F3(k) = k + 1 F7(k) = k + 5k + 6k + 1 3 7 5 3 F4(k) = k + 2k F8(k) = k + 6k + 10k + 4k 4 2 8 6 4 2 F5(k) = k + 3k + 1 F9(k) = k + 7k + 15k + 10k + 1

Table 7.1: Fibonacci polynomials

the Fibonacci numbers are the image under x = 1 of Fn(x) as n → ∞. Next is one of the main

theorems of this section.

th th Proposition 8. Let α = [k] and Fn(x) be the n Fibonacci polynomial. If 1/qn is the n projection

of α, then

qn = Fn(k). (7.9)

Proof. The proposition follows easily by induction.

As a result of Theorem 8, we introduce some notation that helps move things along more

cleanly. When restricted to problems concerning projections of constant continued fractions we

agree to write 1/qn(k) = 1/Fn(k) to keep notation resembling other portions of the article.

Proposition 9. Let α = [k] for k any positive integer. Then Landau’s Theorem 25 and Theorem 26

generalize to the Landau sum of α.

Proof. We use Proposition 6. In two cases, letting u = 2k and u = 4k ± 1. We get a sum of the

form (6.16) under the same conditions that we used to prove Landau’s argument. The proof then

is the same.

We give a table of qn(x) for 2 ≤ n ≤ 9 and 1 ≤ x ≤ 4 to give some impression of the growth of these numbers. We remark that the ﬁrst and second columns are the Fibonacci numbers and Pell numbers, respectively.

42 qn(k) 1 2 3 4 2 1 2 3 4 3 2 5 10 17 4 3 12 33 72 5 5 29 109 305 6 8 70 360 1292 7 13 169 1189 5473 8 21 408 3927 23184 9 34 985 12970 98209

Table 7.2: Values of qn(k)

43 8: LATERAL LANDAU SUMS

This is the ﬁrst section we begin to see analytic techniques playing a role; however, the topics remains ﬁrmly classical. The discussion is followed by two results, neither new, though the angle is presumably novel.

In the previous section we looked at Landau sums or sums of the form

∞ X 1 , given some x ∈ . q (x) N n=0 n

We could change the problem by studying sums of the form

∞ X 1 , given some n ∈ . q (x) N x=1 n

Note, however, that by taking n = 2, then we retrieve the non-converging harmonic sum

∞ ∞ X 1 X 1 = . q (x) x x=1 2 x=1

Therefore, we stipulate that n > 2.

Deﬁnition 22. Recall that qn(x) is actually the Fibonacci polynomial Fn(x). Then For any n > 1, we will call the sum of the form ∞ X 1 (8.1) q (x) x=1 n 44 a lateral Landau sum.

For the ﬁrst of these converging sums i.e. where n = 3 we will mention two proofs of its value, though the proofs aren’t entirely unalike. Both rely on Fourier transform methods.

Proposition 10. ∞ ∞ X 1 X 1 π 1 = = coth π − (8.2) q (x) x2 + 1 2 2 x=1 3 x=1 Proof. See [11], p. 244 and [23], pp. 134-35.

In passing we want to note the similarity of (8.2) to another interesting classical series. In [22].

we see that ∞ ∞ X 1 X = ζ(2n + 2)x2n, for any |x| < 1, (8.3) n2 − x2 n=1 n=0

P∞ p where of course ζ(p) = n=1 1/n .

Proposition 11. Let γ be the number deﬁned in (3.20) and ψ(x) be the derivative of the logarithmic

gamma function Γ(x). In other words, ψ(x) = Γ0(x)/Γ(x). Then

∞ ∞ X 1 X 1 1 √ √ = = 2γ + ψ(i 2) + ψ(−i 2) (8.4) q (x) x3 + 2x 4 n=1 4 x=1

Proof. For this we use the identity found in [18], p. 458. Speciﬁcally, we use

∞ X x ψ(x + 1) = −γ + for x ∈ , x 6= 0, −1, −2, ... . (8.5) k(k + x) C k=1

First substitute the right side of (8.5) for ψ(x) in the right side of (8.4). Then substitute the values

ρ and κ(ρ). Simplify. We ﬁnd the equivalence

∞ ∞ X 1 1 X 1 1 1 = √ − √ . (8.6) k3 + 2k 4 k k=1 k=1 k + i 2 k − i 2 45 P 1/qn(x) 1 2 3 ∞ 1 1/q1(1) 1/q1(2) 1/q1(3) ··· 1.07667 2 1/q2(1) 1/q2(2) 1/q2(3) ··· 0.48415 3 1/q3(1) 1/q3(2) 1/q3(3) ··· 0.25022 ...... ∞ 3.35988 1.84220 1.47679 ··· 0

Table 8.1: A glimpse of the spectrum

We can regroup the right to get

√ ∞ √ ∞ i 2 X 1 −i 2 X 1 = . (8.7) 4 k k2 + 2 k3 + 2k k=1 k=1

This completes the proof.

P P It is only for 1/q3(x) and 1/q4(x) that we have found closed form solutions. But just using the fact that 1 p(x2 + 4)2n+1 = √ √ (8.8) qn(x) (x + x2 + 4)n − (x − x2 + 4)n it is easy to generate Landau sums and lateral Landau sums to relatively low degrees of accuracy.

Finding the Landau sums using ϑ-identities is especially fast; in [5], p 93, it is explained that these numbers are computable in quadratic time. Related to this is also the interesting paper [3].

Experimental calculations show no apparent patterns to these numbers and their behavior as a class is quite interesting. We just scratch the surface in the following table: If C(n) is the sum in the nth column and R(m) the sum in the mth row, consider the region bounded by {(x, y) =

(C(n),R(n)) : for all n ∈ N}. It is clearly a concave region where behavior on the boundary gives a full characterization of arbitrary sums deﬁned by the continuant polynomials.

46 9: SPECULATIONS AND FINAL REMARKS

In this ﬁnal section we are forward-looking. Recall that

n X 1 γ = lim − log(n) . (9.1) n→∞ x x=1

P∞ P∞ R n 1 We know that x=1 1/F2(x) = x=1 1/x and 1 x dx = log(n). Therefore, we consider the analogous limit for n Z n X 1 1 − dx , (9.2) x2 + 1 x2 + 1 x=1 1

2 where we have already seen that 1/(x + 1) = 1/F3(x). With a little work it can be shown that

n Z n X 1 1 1 1 lim − dx = + π − coth(π). (9.3) n→∞ x2 + 1 x2 + 1 2 2 x=1 1

We can speculate about the transcendence of this number. Of course we can formulate this same problem for each n of the sequence 1/Fn(x). We go ahead and make the following conjecture:

Conjecture 1. For any natural number n such that 2 ≤ n,

N Z N X 1 1 lim − dx = ξn (9.4) N→∞ F (x) F (x) x=1 n 1 n is irrational.

47 Other related inquiries deal with the Markov spectrum. We give only a few words; a complete

source is [1]. Recall Hurwitz’s approximation result, Theorem 21. Hurwitz also showed that if we √ √ √ remove (1 + 5)/2 from the set of irrational numbers, then we may substitute 2 2 for 5, and

satisfy the inequality for arbitrary α over our new set; for details, consult [8], p. 20. We make a

conjecture regarding the constant continued fractions.

√ 1+ 5 Conjecture 2. Let α = [k] for some k a natural number. Remove the subset {β : 2 ≤ β ≤ α}

p 2 from the set of irrational numbers. Let C(k) = (k + 1) + 4. Then when pi/qi ,→ α, we have

p 1 α − n+i ≤ (9.5) 2 qn+i C(k)qn+i

holds for i = 0, i = 1 or i = 2.

Next, we give a few more analogues to problems in Classical analysis.

The ζ-function as considered by Euler is deﬁned

∞ X 1 ζ(s) = ns n=1

P∞ 1 π2 for 1 < s ∈ N. In particular, we have the result n=1 n2 = 6 due to Euler. Indeed, Euler found an explicit formula for the numbers ζ(2k) for 0 < k, k an integer, and made use of the following product Y ζ(s) = (1 − p−s)−1. (9.6) p prime

We also know that Riemann studied ζ(s) for s ∈ C such that <(s) < 1. Of course this is the origin of the very famous Riemann Hypothesis.

Since the lateral Landau sums appear a bit more well behaved, it would be interesting to imitate both Euler and Riemann and search for similar identities or behaviors. There are a number of

48 questions we can ask about sums given in Deﬁnition 22, or those of the form

∞ X 1 for chosen n > 1. F (x)s x=1 n

Note that we can take s = 1/2. For what s the sum diverges is not clear to the author, but we speculate for s ≤ 0. A couple of questions we have already answered in Propositions 8.4 and 8.5.

Considering another example, in the easiest case, does an Euler product similar to (9.6) exist for the sum, ∞ X 1 , (9.7) F s n=2 n

th where Fn is the n Fibonacci number? This is doubtful. But we could ﬂip things around and investigate constants of the form ∞ Y −s −1 (1 − Fn ) . (9.8) n=3 What does (9.8) look like, taking reasonable complex numbers s?

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