HURWITZIAN CONTINUED FRACTIONS CONTAINING A REPEATED CONSTANT AND AN ARITHMETIC PROGRESSION

Transcription

1 HURWITZIAN CONTINUED FRACTIONS CONTAINING A REPEATED CONSTANT AND AN ARITHMETIC PROGRESSION Abstract We prove an explicit formula for infinitely many convergents of Hurwitzian continued fractions that repeat several copies of the same constant and elements of one arithmetic progression, in a quasi-periodic fashion The proof involves combinatorics and formal Laurent series Using very little analysis we can express their limits in terms of modified Bessel functions and Fibonacci polynomials The limit formula is a generalization of Lehmer s theorem that implies the continuous fraction expansions of e and tan, and it can also be derived from Lehmer s work using Fibonacci polynomial identities We completely characterize those implementations of our limit formula for which the parameter of each Bessel function is the half of an odd integer, allowing them to be replaced with elementary functions Introduction It is a remarkable property of infinite continued fractions that they often define a sequence whose limit is easier to describe than the individual entries Even for [,, ], the simplest example, the limit is easily found by solving a quadratic equation, it takes a bit longer to find a formula for the convergents, in terms of Fibonacci numbers The class of Hurwitzian continued fractions of the form [α,, α, β }{{} 0, α,, α, β 0 + β n] }{{} n= r d that we propose investigating seems to be no different in this regard For the special case r = 0, D H Lehmer [9] proved a formula for the limit in terms of Bessel functions, that can be verified very easily, after having guessed the correct answer Using another result from [9] and some Fibonacci polynomial identities, it is not hard to generalize Lehmer s formula to Hurwitzian continued fractions of the above form see Remark 20 The resulting generalization has several famous special cases, the most famous ones being Euler s formula for e and the formula for tan On the other hand, the only research regarding the convergents themselves seems to be the work [8] of D N Lehmer D H Lehmer s father!, who proved congruences for their numerators and denominators by induction, in somewhat greater generality, but without stating the values of the convergents in an explicit fashion Our work contains such an explicit formula, stated in Section 2 and proved in Section 4 A variant of the Euler-Mindig formulas using shifted partial denominators yields a summation formula with many vanishing terms This leads to a compact recurrence Date: November, Mathematics Subject Classification Primary 05A0; Secondary 05A5, A55, 30B0, 33A40 Key words and phrases Hurwitzian continued fraction, convergents, Bessel functions, Fibonacci polynomials, Lucas polynomials

2 2 for the numerators of the nd + r st convergents that may be restated as a linear differential equation for a formal Laurent series Classical textbooks on differential equations instruct us to solve the associated homogeneous equation first and then find the general solution by variation of parameters Unfortunately, in our case the solution of the homogeneous equation turns out to be a a two-way infinite formal sum After discarding infinitely many terms to have a formal Laurent series, the spirit of the classical method still inspires a good guess for the form of the solution, where the transformed differential equation encodes a recurrence that is easily solved by inspection Returning to the original Laurent series involves using two polynomial summation formulas that can be shown purely combinatorially, and seem to be interesting by their own right These formulas are shown in Section 3 As outlined in Section 2, our explicit formulas for the nd + r st convergents allow us to calculate the limits using very little analysis To find only the limits, this approach is a bit more tedious than the one outlined in Lehmer s work [9], but we gain a little more insight by also obtaining asymptotic formulas for the numerators and the denominators of the convergents Finally, Section 5 is a reflection on Komatsu s recent remark [7] stating that all known examples of Hurwitzian continued fractions seem to have a short quasi-period and involve hyperbolic trigonometric functions For our class of Hurwitzian continued fractions, the limit may be expressed in terms of modified Bessel functions, which are known to have an elementary form, in terms of rational and hyperbolic trigonometric functions, when their parameter is the half of an odd integer We describe all continued fractions in our class that yield such parameters, and find that, subject to this restriction, the quasi-period d can not be longer than 4 Our work inspires several questions worth exploring in the future Extending the validity of our formulas to all convergents of the same class seems to require only a little more work As already indicated in Lehmer s paper [9], the calculation of the limits is easily extended to the class of Hurwitzian fractions whose quasi-periodic part contains several different constants and one arithmetic sequence An explicit formula for the convergents should be obtained using some multivariate generalization of Fibonacci polynomials The ideas used in our calculations of the convergents may also be useful in finding convergents of other Hurwitzian continued fractions Finally, as indicated in Sections 3 and 4, there is combinatorics behind the formulas for the convergents This combinatorics would be worth uncovering A good starting point may be revisiting the weighted lattice-path model proposed in Flajolet s work [3], where convergents of generalized continued fractions arise as weights of infinite lattice paths of limited height Preliminaries Simple continued fractions A good reference on the basic facts of the subject is Perron s classic work [3, 4] A generalized finite continued fraction is an expression of the form a 0 + a + a 2 + b b 2 b n a n + b n a n,

3 HURWITZIAN CONTINUED FRACTIONS 3 where the initial term a 0, the partial denominators a,, a n and the partial numerators b,, b n may be numbers or functions An infinite continued fraction is obtained by letting n go to infinity The arising questions of convergence have a reassuring answer for simple continued fractions where all b i equal, the initial term a 0 is an integer, and the partial denominators a i for i > 0 are positive integers In the present work we will only use simple continued fractions, and refer to them as continued fractions Every rational number may be written uniquely as a finite continued fraction, subject to the restriction that the last partial denominator is at least 2 [3, 9, Satz 2] Every infinite continued fraction converges to an irrational number and every irrational number may be uniquely written as a necessarily infinite continued fraction [3, 2, Satz 26] As usual, for finite, respectively infinite, continued fraction we will use the shorthand notations [a 0, a,, a n ] and [a 0, a, ], respectively An infinite continued fraction [a 0, a, ] is the limit of its convergents, that is, of the finite continued fractions [a 0,, a n ], obtained by reading the first n partial denominators The convergents [a 0,, a n ] = p n q n 2 may be recursively computed from the initial conditions p =, p 0 = a 0, q = 0 and q 0 =, and from the recurrences p n = a n p n + p n 2 and q n = a n q n + q n 2 for n, cf [3, 2, Eq 2, 3] The integers p n and q n are relative prime for all n [3, 9, Satz 2] The Euler-Mindig formulas, derived for general continued fractions in [3, 3], allow to express the numerators p n and the denominators q n directly Definition We call a set S of integers even if it is the disjoint union of intervals of even cardinality Given two sets of integers S and T such that S T, we say that T evenly contains S, denoted by S e T or T e S, if T \ S is an even set Thus, for example {5} e {,, 7} since the difference {, 2, 3, 4, 6, 7} is the disjoint union of {,, 4} and {6, 7}, whereas {4} e {,, 4} as {, 2, 3} can not be written as the disjoint union of intervals of even cardinality The Euler-Mindig formulas for simple continued fractions may be restated as p n = a i and q n = a i 3 S e{0,,n} i S S e{,,n} i S The following observation is an immediate consequence of Eq 3 Lemma 2 The denominator q n associated to [a 0, a, ] is the same as the numerator p n associated to [a, a 2, ] 2 Hurwitzian continued fractions A Hurwitzian continued fraction is a continued fraction of the form [a 0,, a h, φ 0 λ,, φ k λ] λ=0, where φ 0,, φ k are polynomial functions that send positive integers into positive integers The definition given in [3, 32] is easily seen to be equivalent In the most trivial examples of a Hurwitzian continued fraction all functions φ j are constants, we then obtain a periodic continued fraction The set of real numbers represented by a periodic continued fraction is exactly the set of quadratic irrationals [3, 20] It should be noted that quadratic irrationals are sometimes excluded from the definition of Hurwitzian continued fractions

4 4 in some recent papers [7], by requiring that at least one of the repeatedly used polynomial functions be non-constant A real number is Hurwitzian if its continued fraction representation is a Hurwitzian continued fraction A famous Hurwitzian number is e = [2,, 2,,, 4,,, 6,,, 8,,, 0, ], see [3, 34, Eq 0] The main result on Hurwitzian numbers is Hurwitz theorem [3, 33] stating that for a Hurwitzian number ξ 0 R, and any rational numbers a, b, c, d, satisfying ad bc 0, the number aξ 0 + b/bξ 0 + d is also a Hurwitzian number Hurwitz theorem provides also some estimate on the degrees of the polynomial functions appearing in the continued fraction representation of aξ 0 + b/bξ 0 + d, remains silent however on the issue how the length of the period may be affected by the transformation ξ 0 aξ 0 + b/bξ 0 + d A generalization of Hurwitz theorem may be found in [6] Hurwitzian numbers may be computed from their continued fraction representation in some special cases, the expansion of the set of examples is subject of ongoing research A frequently overlooked first attempt may be found in Perron s book [4, 48, Satz 63] which states without proof a formula for generalized continued fractions of the form having the property that the partial numerators b i are all equal, and that the numbers a i form an arithmetic sequence For simple continued fractions, Perron s result gives [β 0, β 0 + β, β 0 + 2β, β 0 + 3β, ] = β n=0 β 2nn!Γ β0 +n β n=0 β 2nn!Γ β0 +n+ β 4 The same class of Hurwitzian continued fractions was revisited by D H Lehmer [9], who proved the following formula [β 0, β 0 + β, β 0 + 2β, β 0 + 3β, ] = I β0 /β 2β /I β0 /β 2β 5 where I ν z is the modified Bessel function of the first kind z/2 ν+2m I ν z = Γm + Γν + m + 6 D H Lehmer s result was generalized in recent papers to other Hurwitzian continued fractions [5, 7, ] Some analogous results were found for Tasoev continued fractions [5, 6,, 2], defined as continued fractions of the form [a 0 ; a k,, a k As pointed } {{ } m out by T Komatsu [7], most of these recent results involve Hurwitzian and Tasoev continued fractions where the length of the quasi-period does not exceed 4, Komatsu s work [7] contains some sophisticated examples of longer quasi-periods 3 Bessel functions Besides the modified Bessel functions I ν x, our formulas will also involve the original Bessel functions of the first kind m z/2 ν+2m J ν z = Γm + Γν + m + 7 ] k=

5 HURWITZIAN CONTINUED FRACTIONS 5 There is an elementary expression for the functions I ν z and J ν z respectively, whenever ν is the half of an odd integer Indeed, using Γ/2 = π and Γz + = zγz, it is easy to derive directly from the definitions 6 and 7 that we have 2 2 I /2 z = πz coshz, I /2z = sinhz, 8 πz 2 2 J /2 z = πz cosz, and J /2z = sinz 9 πz see [0, List of formulæ: 44, 48, 82, 86] The functions I ν z and J ν z satisfy very similar recurrence formulas: I ν+ x = I ν x 2ν x I νx and J ν+ x = J ν x + 2ν x J νx, see [, 927, 9626] These allow us to find explicit elementary expressions for I ν z and J ν z, whenever ν is the half of an odd integer In particular, for ν = 3/2 we obtain 2 I 3/2 x = coshx sinhx 2 sinx and J 3/2 x = cosx, πx x πx x 0 see [0, List of formulæ: 45, 83] 4 Fibonacci and Lucas polynomials The Fibonacci and Lucas polynomials are q-analogues of the usual Fibonacci and Lucas numbers Definition 3 We define the Fibonacci polynomials F n q and the Lucas polynomials L n q by the initial conditions F 0 q = 0, F q =, L 0 q = 2 and L q = q; and by the common recurrence P n q = q P n q + P n 2 q for n 2, where the letter P should be replaced by either F or L throughout the defining recurrence Fibonacci and Lucas polynomials are widely studied, they even have their own Mathworld and Wikipedia entries As a sample reference on Fibonacci polynomials see Yuang and Zhang [7], for further generalizations, see Cigler [2] The coefficients of the Fibonacci and Lucas polynomials are listed as sequences A02426 and A in the Online Encyclopedia of Integer Sequences [5] In a recent work of Foata and Han [4] the coefficient of q n+ m in F n+ q appears as the number of t-compositions of n into m parts Here we record another combinatorial interpretation The coefficient of q n+ m in F n+ q is the number of m-element even sets contained in {,, n}: F n+ q = q n S = q S es {,,n} S e{,,n} Equation may be shown by induction on n, using the defining recurrence Just like for the usual Fibonacci and Lucas numbers, a closed formula for F n q and L n q may be obtained after solving the characteristic equation We have F n q = q2 + 4 ρn ρ n 2 and 2 L n q = ρ n + ρ n 2 3

6 6 where ρ = q + q and ρ 2 = q q Our main result All our results will be about Hurwitzian continued fractions of the form ξα, β 0, β, d, r := [α,, α, β }{{} 0, α,, α, β 0 + β n }{{} r d where α, β 0, β and d are positive integers and and r is a nonnegative integer In order to state them we need to introduce two magic numbers associated to such a continued fraction ] n= Definition 2 The magic sum associated to ξα, β 0, β, d, r is the sum σα, β 0, β, d, r = β 0 α β + L dα β F d α, the magic quotient associated to ξα, β 0, β, d, r is the quotient ρα, β 0, β, d, r = d β 2 F d α 2 Whenever this does not lead to confusion, we will omit the parameters α, β 0, β, d, r and denote the magic numbers simply by σ and ρ, respectively Note that σ does not depend on r and that ρ depends only on α, β and d Our main result is the following Theorem 22 The nd + r st convergent of ξα, β 0, β, d, r may be written as p nd+r /q nd+r where p nd+r is given by n/2 p nd+r n + σ k = F r+ α ρ k + d r F d r α F d α n β n and q nd+r is given by q nd+r F d α n β n n k! k! n /2 F d αβ n/2 = F r α n k! k! n /2 F d αβ n k! k! n 2k n + σ k n 2k n k! k! n + σ k n 2k ρ k+, ρ k + d+ r F d r α n + σ k n 2k ρ k+ We postpone the proof of Theorem 22 till Section 4 In this section we only show how to find an asymptotic formula for the numerator and the denominator, which yields essentially the same limit formula that can also be derived from Lehmer s work [9] This is the only part of our paper where the notion of limits and convergence from analysis will be used, the proof of Theorem 22 will only involve purely algebraic and combinatorial manipulations We begin with estimating the magic sum σ

7 HURWITZIAN CONTINUED FRACTIONS 7 Lemma 23 The magic sum σ is always positive Proof Since β 0 and β are positive, it suffices to prove that L d α αf d α holds for all positive integer α and all nonnegative integer d This may be shown by induction on d, using the common recurrence of the Fibonacci and Lucas polynomials Corollary 24 The falling factorial σ +n n = σ +n σ is always positive As a consequence, we may divide p nd+r /F d α n β n, as well as q nd+r /F d α n β n by σ + n n and ask whether these quotients converge as n goes to infinity It will turn out that they do Lemma 25 Let σ be any positive real number and ρ be any real number Then n/2 n k! n + σ k lim ρ k ρ m = n σ + n n k! n 2k m!σ + m m Proof Note first that the series on the right hand side is absolute convergent for any ρ as each of its term has smaller absolute value than the corresponding term of e ρ/σ = ρm σ m /m!, an absolute convergent series For any ε > 0 there is an M such that ρ m m!σ + m m < ε 3 2 m=m+ Let us consider now the expression of n on the left hand side It may be rewritten as n/2 ρ k k!σ + k k δ n,k where δ n,k = n k k σ + n k The factor δ n,k may be estimated as follows: k n 2k + δ n,k = σ + n k n k n 2k + σ + n σ + n k k n k σ + n Our upper bound for δ n,k is for k = 0 and it is less than for k > 0, since n + σ > n > n k For fixed k, our lower bound converges to as n goes to infinity Now, for all n > M we may write n/2 ρ k k!σ + k k δ n,k = M ρ k k!σ + k k δ n,k + n/2 k=m+ ρ k k!σ + k k δ n,k By our choice of M, and by δ n,k, the second sum on the right hand side has absolute value less than ε/3, for all n > M As n goes to infinity, each of δ n,0,, δ n,m converges to, so there is an n 0 such that for all n > n 0 we have implying M ρ k k!σ + k k δ n,k n/2 ρ k δ n,k k!σ + k k M ρ k k!σ + k k < ε 3, M ρ k k!σ + k k < 2ε 3, 22

8 8 Combining 2 and 22 we obtain n/2 ρ k ρ k δ n,k k!σ + k k k!σ + k k < ε for all n n 0 Lemma 26 Let σ be any positive real number and ρ be any real number Then n /2 n k! n + σ k lim ρ k+ ρ m+ = n σ + n n k! n 2k m!σ + m m+ Proof The proof is very similar to the proof of Lemma 25, thus we omit the details We only note that this time we have n /2 n /2 n k! n + σ k ρ k+ ρ k+ = δ n,k σ + n n k! n 2k k!σ + k k+ where δ n,k = n k k σ + n k Again δ n,k for all n and k and, for any fixed k, the limit of δ n,k is as n goes to infinity Combining Lemmas 25 and 26 with Theorem 22 we obtain p nd+r ρ m lim = F n F d α n β n r+ α σ + n n m!σ + m m and lim n q nd+r F d α n β n σ + n n = F r α + d r F d r αf d αβ ρ m m!σ + m m ρ m+ m!σ + m m d r+ ρ m+ F d r αf d αβ m!σ + m m+ 24 Taking the quotients of the right hand sides of 23 and 24 we obtain the following formula for ξ = ξα, β 0, β, d, r: ξ = F r+ α F r α ρ m m!σ + m m + d r F d r αf d αβ ρ m m!σ + m m + d r+ F d r αf d αβ ρ m+ m!σ + m m+ ρ m+ m!σ + m m+ 25

9 HURWITZIAN CONTINUED FRACTIONS 9 This is the formula that could have been discovered in a parallel universe where interest in Hurwitzian continued fractions were to arise a long time before defining Bessel functions We may restate Eq 25 in terms of Bessel functions using the following two obvious statements Lemma 27 For any positive ρ and σ we have ρ m = I σ 2 ργσ and m!σ + m m ρ σ /2 Here I ν z is the modified Bessel function defined in 6 Lemma 28 For any negative ρ and positive σ we have ρ m = J σ 2 ργσ m!σ + m m ρ σ /2 ρ m+ = J σ2 m!σ + m m+ Here J ν z is the Bessel function given by 7 ρ m+ = I σ2 m!σ + m m+ ργσ ρ σ 2/2 and Using Lemmas 28 and 27 above we may rephrase Eq 25 as follows ργσ ρ σ 2/2 Theorem 29 Let α, β 0, β and d be positive integers, and let r be nonnegative integers Then the Hurwitzian continued fraction is given by ξα, β 0, β, d, r = [α,, α, β }{{} 0, α,, α, β 0 + β n }{{} r d ] n= ξα, β 0, β, d, r = F r+αi σ 2 ρ + r+ F d r αi σ 2 ρ F r αi σ 2 ρ + r F d r αi σ 2 ρ if d is odd, and it is given by ξα, β 0, β, d, r = F r+αj σ 2 ρ + r+ F d r αj σ 2 ρ F r αj σ 2 ρ + r F d r αj σ 2, ρ if d is even Here I ν z and J ν z, respectively, denotes the is the modified respectively original Bessel function defined in 6 and 7, respectively Proof We work out only the case of odd d in detail, the case of even d is completely analogous Direct substitution of Lemma 27 into 25 yields F r+ α I σ 2 ρ + d r I σ 2 ρ F ρ ξα, β 0, β, d, r = σ /2 d r αf d αβ ρ σ 2/2 F r α I σ 2 ρ + d r+ I σ 2 ρ F ρ σ /2 d r αf d αβ ρ σ 2/2 After multiplying the numerator and the denominator by ρ σ /2 we get ξα, β 0, β, d, r = F r+αi σ 2 ρ + d r ρf d r αf d αβ I σ 2 ρ F r αi σ 2 ρ + d r+ ρf d r αf d αβ I σ 2 ρ

10 0 The first equality in Theorem 29 follows from ρ = β F d α and from the fact that d = in this case A similar reasoning for even d yields ξα, β 0, β, d, r = F r+αj σ 2 ρ d r F d r αj σ 2 ρ F r αj σ 2 ρ d r+ F d r αj σ 2 ρ, a final simplification may be made by observing that d = in this case Remark 20 Theorem 29 may also be derived from Lehmer s work [9] directly, using a few, easily verifiable facts about Fibonacci and Lucas polynomials Again, we outline the proof for odd d only, the case of even d being completely analogous Consider first the case when d = and r = 0 In this case we get σ = β 0 /β, regardless of of α Theorem 29 takes the form ξα, β 0, β,, 0 = I σ 2 ρ/i σ 2 ρ, which, by ρ = β 2, is exactly Lehmer s formula 5 Consider next the case when d is an arbitrary odd integer and r = 0 In this case, we may use Lehmer s [9, Theorem 4] with all constants being equal to α Using the fact that [α,, α] = F d+α }{{} F d α, d the fractions A/B, A /B and A /B appearing in [9, Theorem 4] are easily seen to correspond to β 0 αf d α+f d+ α/f d α, β 0 αf d α+f d α/f d α and F d α/f d α, respectively, in our notation Lehmer s bb+b +B /ab corresponds to our β 0 F d α + 2F d α β F d α = β 0 αf d α + αf d α + 2F d α β F d α = σ, since αf d α + 2F d α = L d α holds for all d Lehmer s [9, Theorem 4] gives ξα, β 0, β, d, 0 = F d α + I σ 2β F d α F d α I σ 2β F d F d α which is the same as the formula implied by Theorem 29, after noting that we have ρ = β F d α 2 Finally, for arbitrary r we may substitute η := ξα, β 0, β, d, 0 into the formula [α,, α, η] = F r+αη + F r α }{{} F r αη + F r α r and obtain ξα, β 0, β, d, r = F r+ α F r α F d α F d α Recall that σ does not depend on r Equivalently, F d α + I σ 2β F d α I σ2β F d F d + F α r α F d α + I σ 2β F d α I σ2β F d F d + F α r α ξ = F r+αi σ 2β F d α + F r αf d α F r+ αf d α I σ 2β F d F d α F r αi σ 2β F d α + F r αf d α F r αf d α I σ 2β F d F d α Theorem 29 now follows after observing that F r αf d α F r+ αf d α and and F r αf d α F r αf d α, respectively, may be replaced by r+ F d r α and r F d r α respectively

11 HURWITZIAN CONTINUED FRACTIONS 3 Two useful lemmas In this section we provide a combinatorial proof for two polynomial identities which seem to be interesting by their own right They will play a crucial role in Section 4 where we prove Theorem 22 Our lemmas will be summation formulas for the polynomials and S n x, y := n! R n x, y := n! n n k Lemma 3 The polynomials R n x, y satisfy n x n m n m! R mx, y = n/2 n n! k n + y n k n k! 2 x k 3 n n n + y n k!n k!x k+ 32 k n k n + y k n 2k n k! 2 x k for all n 0 Proof We consider both sides of the equation as polynomials in the variable y with coefficients from the field Qx Since a nonzero polynomial has only finitely many roots, it suffices to show that the two sides equal for any nonnegative integer value of y For y N, the left hand side may then be rewritten as n n m m m + y x n m m k! 2 x k n! n m k y + k This is /n! times the total weight of all quadruplets π, π 2, γ, γ 2 subject to the following conditions: π is a permutation of {,, n}, π 2 is a permutation of {,, n + y}, the functions γ : {,, n} {0,, 2} and γ 2 : {,, n + y} {0,, 2} are colorings; 2 for i {, 2} an element j satisfying γ i j = must be a fixed point of π i ; 3 for any j, γ j = 2 is equivalent to γ 2 j = 2 and γ j = γ 2 j = 2 implies that j is a common fixed point of π and π 2 ; 4 the colorings γ and γ 2 satisfy {j {,, n + y} : γ 2 j > 0} = {j {,, n} : γ j > 0} + y; We define the weight of π, π 2, γ, γ 2 as x {j {,,n} : γ j=} x {j {,,n} : γ j=2} Indeed, after setting n m as the number of elements j satisfying γ j = γ 2 j = 2, there are n n m ways to select them By rule 3 these are common fixed points of π and π 2 and thus elements of {,, n} Next we set k as the number of elements j {,, n} satisfying γ j =, and select these elements, in m k ways The remaining elements of {,, n} satisfy γ j = 0 At this point we have {j {,, n} : γ j > 0} = n m + k Thus, by rule 4, the set {j {,, n + y} : γ 2 j > 0} must have n m + k + y elements Exactly n m of these elements satisfy γ 2 j = 2, therefore the number of elements j {,, n + y} satisfying γ 2 j = must be y + k

12 2 There are m+y y+k ways to select them So far we have selected n m + k fixed points of π and n m + y + k fixed points of π 2 We can complete prescribing the action of π and π 2 in m k! 2 ways The same total weight may also be found by fixing the pair π, π 2 first, and summing over all allowable pairs of colorings γ, γ 2 We call the pair γ, γ 2 allowable, if the quadruplet π, π 2, γ, γ 2 satisfies the conditions listed above Suppose j 0 is a common fixed point of π and π 2 Observe that the contribution of all allowable pairs γ, γ 2 satisfying γ j 0 = γ 2 j 0 = 2 cancels the contribution of all allowable pairs satisfying γ j 0 = γ 2 j 0 = Indeed, let γ, γ 2 an allowable pair satisfying γ j 0 = γ 2 j 0 > 0 Then the pair γ, γ 2 defined by { γ ij γ i j if j j 0 = 3 γ i j if j = j 0 is also allowable and also satisfies γ j 0 = γ 2j 0 > 0 Here j {,, n} for γ and γ and j {,, n + y} for γ 2 and γ 2 The map γ, γ 2 γ, γ 2 is an involution that matches canceling terms: the only difference between the respective contribution is a factor of x or x associated to j 0 Therefore we may restrict our attention to the total weight of quadruplets π, π 2, γ, γ 2 satisfying the following additional criteria: 5 no j satisfies γ j = γ 2 j = 2, in particular, the colorings γ and γ 2 map into the set {0, }; 6 no j satisfies γ j = γ 2 j =, in other words, the sets {j {,, n} : γ j = } and {j {,, n + y} : γ 2 j = } are disjoint When computing the total weight of such quadruplets, we may first select k as the number of elements j {,, n} satisfying γ j = and select them in n k ways By rule 4 there must be y + k elements j {,, n + y} satisfying γ 2 j = and, by rule 6, this set is disjoint of the previously selected k-element set Thus there are n+y k y+k ways to select them So far we have selected k fixed points of π and y +k fixed points of π 2 There are n k! 2 ways to complete prescribing π and π 2, the weight of the quadruplet is x k We obtain a total contribution of n n n+y k k y+k n k! 2 x k which is exactly n! times the right and side Lemma 32 The polynomials S n x, y satisfy n for all n 0 x n m n m! S mx, y = n! n /2 n n + y k n k!n k!x k+ k n 2k Proof A proof may be obtained by performing slight modifications to the proof of Lemma 3, which we outline below The left hand side equals x n n m m m + y x n m m k!m k!x k, n! n m k y + k + which may be considered as x/n! times the total weight of quadruplets π, π 2, γ, γ 2, where the only change to the definition is that, instead of 4, now we require

13 HURWITZIAN CONTINUED FRACTIONS 3 4 the colorings γ and γ 2 satisfy {j {,, n + y} : γ 2 j > 0} = {j {,, n} : γ j > 0} + y + Thus we will have to select y + k + elements j instead of y + k satisfying π 2 j =, in m+y y+k+ In the last stage, we will have selected n m + y + k + fixed points of π2, thus we will only have m k! ways to complete the selection of π 2 The same involution as before shows that we may again restrict our attention to those quadruplets which satisfy the additional conditions 5 and 6 Let us set k again as the number of elements j {,, n} satisfying γ j = The only adjustment we need to make to the reasoning is to observe that now we need to have y +k + elements satisfying γ 2 j = which may be selected in n+y k y+k+ ways, instead of n+y k y+k ways Finally, we may complete the selection of π 2 in n k! ways, instead of n k! The resulting total weight is exactly x/n! times the right hand side of our stated equality For the sake of use in Section 4 let us note that R n x, y and S n x, y may also be written in the following shorter form, using falling factorials: n x k y + n n k n x k+ y + n n k R n x, y = and S n x, y = 33 k! k! A similar simplification yields that Lemma 3 is equivalent to n x n m n/2 n m! R n k! n + y k mx, y = x k for all n 0, 34 k! n 2k and that Lemma 32 has the compact form n x n m n /2 n m! S n k! n + y k mx, y = x k+ for all n 0 35 k! n 2k 4 Calculating the convergents In this section we calculate nd + r st convergent of ξα, β 0, β, d, r directly, from the Euler-Mindig formulas 3 We begin by observing that, for an arbitrary continued fraction [a 0, a, ] and an arbitrary positive integer α, the numerator p n in 3 may be rewritten as p n = a i α + α = a i αα S\T S e{0,,n} i S Changing the order of summation gives p n = a i α T {0,,n} i T S e{0,,n} T S i T T S e{0,,n} α S 4 For ξα, β 0, β, d, r we have a i = α unless i is congruent to r modulo d Thus, to compute p nd+r using 4, we only need to sum over subsets T whose elements are all congruent to r modulo d This observation yields the following recurrence: n p nd+r = F nd+r+ α + p kd+r β 0 + β k α F n kd α 42

14 4 By Eq, the term F nd+r+ α above is the contribution of T =, whereas the term p kd+r β 0 + β k α F n kd+ α is the total contribution of all sets T whose largest element is kd + r Substituting n = 0 into 42 yields the initial condition p r = F r+ α which is obviously true for r > 0, and it is also valid when r = 0 after setting p =, as usual Consider the formal Laurent series yt := p nd+r t β 0+β n α Qt n=0 For yt, Eq 42 yields yt := F nd+r+ αt β n t β0 α + t n=0 F nd αt β n y t, 43 where y t is the formal derivative of yt with respect to t To write 43 in a more explicit form, observe that, by Eq 2, we have F nd+r+ αt n ρ nd+r+ ρ nd+r+ 2 = t n ρ r+ = α2 + 4 α2 + 4 ρ d t ρr+ 2 ρ d 2t n=0 n=0 n=0 Using the fact that ρ ρ 2 =, the above equation may be rewritten as F nd+r+ αt n ρ r+ ρ r+ 2 ρ r+ ρ d 2 ρ r+ 2 ρ d = t α2 + 4 ρ d + ρ d 2t + d t 2 = n= ρ r+ ρ r+ 2 + r+ ρ d r ρ d r 2 t α2 + 4 ρ d + ρ d 2t + d t 2 By Eqs and 3 we obtain F nd+r+ αt n = F r+α + r+ F d r αt 44 L d αt + d t 2 n=0 Note that substituting r = d in Eq 44 yields F nd αt n = t F nd+d αt n F d α t = L d αt + d t, 2 n= n=0 since F 0 α = 0 Using these last two equations, we may rewrite 43 as yt = F r+α + r+ F d r αt β L d αt β + d t 2β t β 0 α + Rearranging to express the derivative of yt yields F d α t β + L d αt β + d t 2β y t y t = yt t β L d αt + d t β F d α F r+α + r+ F d r αt β t β 0 α F d α t β + 45

15 HURWITZIAN CONTINUED FRACTIONS 5 Inspired by the way we solve ordinary differential equations in analysis, we will guess the solution of 45 by solving first the corresponding homogeneous equation and then replace the arbitrary constant by a formal Laurent series The next few lines will not make sense, they just indicate how one may come up with a good guess for yt A reader who does not like obscure reasoning, should skip ahead to 46 and accept that there is no rational explanation as to why introducing the formal Laurent series zt is a good idea The homogeneous equation may be rewritten as y Ht = y Ht t β L d αt + d t β F d α d dt lny Ht = t β L d αt + d t β, F d α yielding y H t = C t L dα/f d t α β + d t β exp β F d α where C is an arbitrary constant This solution to the homogeneous equation suggests looking for a solution to Eq 45 of the form yt = zt t L dα/f d t α β + d t β exp β F d α Equivalently, we would want to set zt := yt t L dα/f d t α β + d t β exp β F d α Alas, the resulting formal expression would contain arbitrary large positive and as well as arbitrary small negative powers of t, it does not resemble a Laurent series at all Hoping that a slight change would not upset our calculations irreparably, we define zt by setting zt := yt t L dα/f d α d t β exp 46 β F d α Note that zt is a formal Laurent series in the variable t /Fdα, an infinite formal sum of the form zt = s n t β n+β 0 α+l d α/f d α Qt /Fdα n=0 Taking the derivative on both sides of 46 yields z t = y t t L d α F d α d t β exp β F d α +yt t L d α F d α d t β Ld αt d t β exp β F d α F d α = t L d α F d α d t β exp y t + yt Ldαt d t β β F d α F d α

16 6 Note that L d α/f d α is always positive After substituting the value of y t from 45 and simplifying we obtain z t = t L d α F d α d t β exp yt F r+α + r+ F d r αt β t β 0 α β F d α F d αt β + By 46, the last equation is equivalent to z t = zt F d αt β + t L d α F d α exp d t β β F d α F r+α + r+ F d r αt β t β 0 α F d αt β + Comparing the coefficients of t β n+β 0 α+l d α/f d α on both sides yields s n β n + β 0 α + L dα F d α = s n+ F d α F r+α F d α + r F d r α F d α d n+ β F d α n +! d n β F d α Note that the factor β n+β 0 α+l d α/f d α in the last equation equals β n+σ, where σ is the magic sum, and that the magic quotient ρ appears twice on the right hand side Since, by Corollary 24, σ + n n+ is not zero, we may divide both sides by F d α n β n+ σ + n n+, and obtain the following recurrence for s n := s n /F d α n β n σ + n n : s n+ = s n + F r+ α ρ n+ r F d r α ρ n n +!σ + n n+ F d αβ n!σ + n n+ Considering the fact that s 0 = s 0 = F r+ α and that r F d r α/f d αβ = d r F d r αf d αβ ρ, the last recurrence implies s n = F r+ α n ρ k n + d r ρ k+ F d r αf d αβ k!σ + k k k!σ + k k+ Multiplying both sides by σ + n n+ yields s n n ρ k σ + n n k = F F d α n β n r+ α k! + d r F d r αf d αβ n n! ρ k+ σ + n n k k! By the formulas given in Eq 33, we may rewrite the previous equation in terms of the polynomials R n x, y and S n x, y as follows: s n F d α n β n = F r+ αr n ρ, σ + d r F d r αf d αβ S n ρ, σ 47

17 By 46 we have HURWITZIAN CONTINUED FRACTIONS 7 yt = zt t L dα/f d α exp d t β β F d α Comparing the coefficients of t β 0+β n α yields d n m n F d αβ p nd+r = s m, n m! hence we have p nd+r F d α n β n = n s m F d α m β m d n m F d α 2 β 2 n m! = n s m F d α m β m ρ n m n m! Substituting Eq 47 into this last equation and using Eqs 34 and 35 yields the formula for p nd+r in Theorem 22 For positive r, the formula for q nd+r stated in Theorem 22 is an easy consequence of the formula for p nd+r Indeed, by Lemma 2, the denominator q nd+r is the same as the numerator p nd+r 2 associated to ξα, β 0, β, d, r, thus we only need to replace r by r in the formula stated for p nd+r 2 It only remains to show the following lemma Lemma 4 The equation stated for q nd+r in Theorem 22 remains valid when we substitute r = 0 Proof Since F 0 α = 0, substituting r = 0 in Theorem 22 gives q nd F d α n β n n /2 = d+ F d α 2 β n k! k! n + σ k ρ k+ n 2k Here we may replace d+ F d α 2 β by ρ β After rearranging we obtain n /2 q nd = F d α n β n n k! n + σ k ρ k 48 k! n 2k We need to show the validity of this equation Observe that ξα, β 0, β, d, 0 = [β 0, α,, α, β }{{} 0 + β, ] = β 0 α + [α,, α, β }{{} 0 + β, ] d d = β 0 α + ξα, β 0 + β, β, d, d Thus the denominator q nd associated to ξα, β 0, β, d, 0 is the same as the same as the denominator qn d associated to ξα, β 0 + β, β, d, d We may apply the already shown part of Theorem 22 Since the magic quotient ρ depends only on α, β and d, it is the same for ξα, β 0, β, d, 0 and for ξα, β 0 + β, β, d, d For the magic sums we get σα, β 0 + β, β, d, d = β 0 + β α β + L dα β F d α = σα, β 0, β, d, 0 +

18 8 Therefore we may obtain an equation for the q nd associated to ξα, β 0, β, d, 0 by replacing n with n, r with d and σ with σ+ in the formula for q nd in Theorem 22 Since F 0 α = 0, the second sum vanishes and we get q nd+r α, β 0, β, d, 0 F d α n β n which is obviously equivalent to 48 n /2 = F d α n k! k! n + σ k ρ k, n 2k 5 Special cases leading to elementary expressions In this section we describe all instances of Theorem 29 for which the magic number σ is the half of an odd integer, forcing all modified Bessel functions in the statement to be known elementary functions There is not much to say about the case d = : as seen in Remark 20, we have σ = β 0 /β thus σ depends only on β 0 and β in a very simple fashion In the case when d 2 we give a similarly simple description Theorem 5 If d 2 then σ is the half of an odd integer if and only if one of the following conditions holds: d = 3, α = and β 0 + /β is the half of an odd integer; 2 d = 2, α = and β 0 + 2/β is the half of an odd integer; 3 d = 2, α = 2 and β 0 + /β is the half of an odd integer; 4 d = 2, α = 4 and 2β 0 + /β is an integer Proof First we show that even assuming that σ is the half of an even or odd integer implies that d is at most 3 The fact that σ belongs to /2 Z implies the same for β σ + α β 0 = L d α/f d α We may subtract any integer multiple of F d α from L d α and still have element of /2 Z Let us select r d α := L d α αf d α and consider the fraction r d α/f d α /2 Z We claim that 0 < r dα F d α < holds for α 2 and d 3 5 Indeed, for d = 3, we have r 3 α = L 3 α αf 3 α = α 3 + 3α αα 2 + = 2α and F 3 α = α 2 + Both r 3 α and F 3 α are positive and r 3 α < F 3 α follows from F 3 α r 3 α = α 2 > 0 for α 2 For d = 4 we have r 4 α = L 4 α αf 4 α = α 4 + 4α αα 3 + 2α = 2α and F 4 α = α 3 + 2α Both r 4 α and F 4 α are positive and r 4 α < F 4 α follows from F 4 α r 4 α = α 3 2α 2 + 2α 2 = α 2 α + 2α > 0 for α 2 For larger values of d, we may show that 0 < r d α < F d α holds when d 3 and α 2 by induction on d, using the fact that the statement is valid for d {3, 4} and that r d α satisfies the same recurrence as F d α, allowing to express the inequality for the next value of d as a positive combination of the inequalities for the current and the previous values of d This concludes the proof of 5 As a consequence of the inequality 5, whenever d 3 and α 2, the only way for r d α/f d α to be an integer is to have r d α/f d α = /2 In other words, in this

19 HURWITZIAN CONTINUED FRACTIONS 9 case, we must have 2r d α = F d α which is equivalent to 2L d α = 2α + F d α To show that α 2 and d 3 can not hold simultaneously, it suffices to show that 2L d α can never be equal to 2α + F d α To do so, first we observe that 2L d α > 2α + F d α holds for α {2, 3} and d 3 Indeed, for α = 2 we have 28 = 2L 3 2 > F 3 2 = 25 and 68 = 2L 4 2 > F 4 2 = 60 and we may prove the same inequality for higher values of d by induction Similarly, for α = 3 we have 72 = 2L 3 3 > F 3 3 = 70 and 238 = 2L 4 3 > F 4 3 = 23 and we may proceed again by induction on d Finally, to exclude α 4, we will show For d = 3 we have and for d = 4 we have 2L d α < 2α + F d α holds for α 4 and d 3 2α + F 3 α 2L 3 α = α 2 4α + = αα 4 + > 0, 2α + F 4 α 2L 4 α = α 3 4α 2 + 2α 4 = α 2 α 4 + 2αα 2 > 0, and again we may proceed by induction on d We obtained that, for d 3, σ /2 Z is only possible if α = In that case, for d 4 we have L d 2F d = F d 3 this may be shown by induction Clearly L d /F d is the half of an integer, if and only if the same holds for F d 3 /F d Now we may exclude the case d 4 completely, after observing that 0 < F d 3 F d < 2 holds for d 4 Indeed, the above inequality is equivalent to 0 < 2F d 3 < F d which may be easily shown by induction We have shown that σ can only be the half of an integer if d = 2 or d = 3 In the case, when d = 3, we have also shown that only α = is possible, and we get σ = β 0 β + L 3 β F 3 = β 0 β + 4 2β = β 0 + β Consider finally the case when d = 2 As before, σ /2Z implies that L d 2/F d 2 = α 2 + 2/α /2Z This implies that α must be a divisor of 4, that is, an element of {, 2, 4} We have σ = β 0 α + L 2α β β F 2 α = αβ 0 α 2 + α2 + 2 β α β α = αβ β α Therefore, for α = we get σ = β 0 + 2/β, for α = 2 we get σ = β 0 + /β and for α = 4 we get σ = 2β 0 + /2β A nice example of the case when d = 3 and α = in Theorem 5 above is the case when α =, β 0 = 3m, β = 2m, d = 3, r = 2 for some m > 0 In this case we get σ = 3/2 and ρ = /6m 2 Theorem 29 gives ξ, 3m, 2m, 3, 2 = 2I /2 /2m I /2 /2m + I 3/2 /2m

20 20 Using 8 and 0 we may rewrite the preceding equation as [,, 3m + 2mn] n=0 = For m = we obtain 2 sinh/2m cosh/2m 2m sinh/2m [,, 2,,, 4, ] = e/2 e /2 = e e /2 A similarly nice example for the case when d = 2 and α = in Theorem 5 above is ξ, 3m 2, 2m, 2, = [, 3m 2 + 2mn] n=0 for some m > 0 In this example σ = 3/2 and ρ = /4m 2 hold Theorem 29 gives ξ = J /2 2 ρ J /2 2 ρ J 3/2 2 ρ = Using 9 and 0 we may rewrite ξ above as ξ = J /2 /m J /2 /m J 3/2 /m sin/m cos/m m sin/m Substituting m = yields tan = [,,, 3,, 5,, 7, ] We conclude this section with an example which is not likely to be found in the literature, due to its sheer ugliness Let us set α = 4, β 0 = 7m + 3, β = 2m +, d = 2 and r =, where m is any nonnegative integer For this example we have σ = 7/2 and ρ = /62m + 2, yielding 2 ρ = /4m + 2 Using the fact that 2 3 J 5/2 z = πz z sinz 3z cosz 2 and J 7/2 z = 2 5 πz z 6 5 sinz 3 z z cosz 2 see [0, List of formulæ: 46, 47], Theorem 29 gives 4 22m + 2 sin 4m+2 62m + cos 4m+2 ξ = 240m m + 53 cos 4m+2 960m m m + 97 sin 4m+2 The above formula was calculated from Theorem 29 with the help of Maple The same program was used to double-check its correctness for selected values of m For example, for m = 0 we obtain 4 sin 2 6 cos 2 53 cos 2 97 sin = [4, 3, 4, 4, 4, 5, 4, 6, 4, 7, 4, ] 2 Acknowledgments This work was partially supported by a grant from the Simons Foundation #24553 to Gábor Hetyei

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