SYMMETRY AND SPECIALIZABILITY IN THE CONTINUED FRACTION EXPANSIONS OF SOME INFINITE PRODUCTS

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1 SYMMETRY AND SPECIALIZABILITY IN THE CONTINUED FRACTION EXPANSIONS OF SOME INFINITE PRODUCTS J MC LAUGHLIN Abstract Let fx Z[x] Set f 0x = x and for n 1 define f nx = ff n 1x We describe several infinite families of polynomials for which the infinite product f nx n=0 has a specializable continued fraction expansion of the form S = [1; a 1x a 2x a 3x ] where a ix Z[x] for i 1 When the infinite product and the continued fraction are specialized by letting x take integral values we get infinite classes of real numbers whose regular continued fraction expansion is predictable We also show that under some simple conditions all the real numbers produced by this specialization are transcendental We also show for any integer k 2 that there are classes of polynomials fx k for which the regular continued fraction expansion of the product k n=0 n= f nx k is specializable but the regular continued fraction expansion of k f nx k is not specializable 1 Introduction The problem of finding the regular continued fraction expansion of an irrational quantity expressed in some other form has a long history but until the 1970 s not many examples of such continued fraction expansions were known Apart from the quadratic irrationals and numbers like e q for certain rational q there were very few examples of irrational numbers with predictable patterns in their sequence of partial quotients 1991 Mathematics Subject Classification Primary:11A55 Key words and phrases Continued Fractions 1

2 2 J MC LAUGHLIN Being able to predict a pattern in the regular continued fraction expansion of an irrational number is not only interesting in its own right but if one can also derive sufficient information about the convergents it is then sometimes possible to prove that the number is transcendental In [10] Lehmer showed that certain quotients of modified Bessel functions evaluated at various rationals had continued fraction expansions in which the partial quotients lay in arithmetic progressions He also showed that similar quotients of modified Bessel functions evaluated at the square root of a positive integer had continued fraction expansions in which the sequence of partial quotients consisted of interlaced arithmetic progressions An old result originally due to Böhmer [3] and Mahler [11] was rediscovered by Davison [7] and Adams and Davison [1] generalizing Davison s previous result in [7] In this latter paper the authors were able to determine for any positive integer a 2 and any positive irrational number α the regular continued fraction expansion of the number 11 S a α = a 1 r=1 1 a rα in terms of the convergents in the continued fraction expansion of α 1 They were further able to show that all such numbers S a α are transcendental A generalization of Davison s result from [7] was given by Bowman in [5] and Borwein and Borwein [4] gave a two-variable generalization of 11 but the continued fraction expansion in this latter case is not usually regular Shallit [15] and Kmo sek [8] showed independently that the continued fraction expansions of the irrational numbers 1 u 2k k=0 have predictable continued fraction expansions This result was subsequently generalized by Köhler [9] by Pethö [13] and by Shallit [16] once again In [12] Mendès France and van der Poorten considered infinite products of the form 1 + X λ h h=0 where 0 < λ 1 < λ 2 < is any sequence of rational integers satisfying a certain growth condition and showed that such products had a predictable continued fraction expansion in which all the partial quotients were polynomials in Z[X] They further showed that if the infinite product and continued fraction were specialized by letting X be any integer g 2 that all such real numbers γ = 1 + g λ h h=0

3 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 3 so obtained were transcendental Similar investigations in which the continued fraction expansions of certain formal Laurent series are determined can be found in [19] [18] [20] and [2] Let fx Z[x] f 0 x = x and for i 1 f i x = ff i 1 x the i-th iterate of fx In [17] Tamura investigated infinite series of the form θx : f = m=0 1 f 0 xf 1 x f m x He showed for all polynomials in a certain congruence class that the continued fraction expansion of θx : f had all partial quotients in Z[x] He further showed that if the series and continued fraction were specialized to a sufficiently large integer depending on fx then the resulting number was transcendental The infinite series k=0 1/x2k investigated by Shallit [15] and Kmo sek [8] may be regarded as a special case of the infinite series k=0 1/f kx with fx = x 2 In a very interesting paper [6] Cohn gave a complete classification of all those polynomials fx Z[x] for which the series k=0 1/f kx had a continued fraction expansion in which all partial quotients were in Z[x] By then letting x take integral values he was able to derive expansions such as the following: n 0 1 = [0; T 4 n ] where T l x denotes the l-th Chebyshev polynomial and also to derive the continued fraction expansion for certain sums of series At the end of Cohn s paper he listed a number of open questions and conjectures One of the problems he mentioned was finding a similar classification of all those polynomials fx Z[x] for which the regular continued fraction expansion of the infinite product f k x k=0 has all partial quotients in Z[x] This turns out to be a technically more difficult problem One reason is that given any positive integer k there are classes of polynomials such as fx k = 2x + x 2 + x k 1 k xgx for which the regular continued fraction expansion of the product k n= /f nx k is specializable for all polynomials gx 1 k+1 mod x but the regular continued fraction expansion of k+1 n= /f nx k is not specializable This is in contrast to the infinite series case dealt with by Cohn where k=0 1/f kx had a specializable continued fraction expansion if and only if 3 k=0 1/f kx had a specializable continued fraction expansion

4 4 J MC LAUGHLIN In this paper we give several infinite classes of polynomials for which n= /f nx has a specializable regular continued fraction For the polynomials in these classes of degree at least three we specialize the product at 12 by letting x take positive integral values producing certain classes of real numbers We examine the corresponding regular continued fractions to prove the transcendence of these numbers 2 Some Preliminary Lemmas Unless otherwise stated fx Gx gx will denote polynomials in Z[x] f 0 x := x and for n 0 f n+1 x := ff n x Sometimes for clarity and if there is no danger of ambiguity fx will be written as f and f n x as f n Likewise fx m will be written as f m f n x m as fn m etc For a fixed fx Z[x] set fx = f = n := 1 + 1fi n n n and i=0 fx = f = := 1 + 1fi Similarly S n fx = S n f = S n will denote the regular continued fraction expansion via the Euclidean algorithm of n and S fx = S f = S will denote the regular continued fraction expansion of The more concise forms will be used when there is no danger of ambiguity Unless stated otherwise the sequence of partial quotients in S n will be denoted by w n so that S n = [ w n ] If a partial quotient in a continued fraction is a polynomial in Z[x] it is said to be specializable A continued fraction all of whose partial quotients are specializable is also called specializable We say that a continued fraction [a 0 a 1 a n ] has even resp odd length if n is even resp odd Since a form of the folding lemma will be used later we state and prove this for the sake of completeness In what follows let w denote the word a 1 a n w the word a n a 1 and w the word a n a 1 For i 0 let A i /B i denote the i-th convergent of the continued fraction [a 0 a 1 ] Recall that 21 and i=0 A n+1 = a n+1 A n + A n 1 B n+1 = a n+1 B n + B n 1 22 A n B n 1 A n 1 B n = 1 n 1 We need the following preliminary results

5 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 5 Lemma 1 For j = [ 1 j w] = 1 j B n B n 1 If a 0 = 1 then 24 [ 1 j w] = 1 j B n A n B n and 25 [ 1 j w 1 j ] = 1 j A n A n 1 Proof All of these follow easily from the correspondence between matrices and continued fractions easily proved by induction or see [22]: a0 1 a1 1 an 1 An A = n B n B n 1 and a0 1 a1 1 an 1 = n An A n 1 B n B n 1 Lemma 2 [19] [a 0 ; w Y w] = A n 1 + 1n B n Y A n B n Proof If we use 23 followed by 21 and then 22 we get that [a 0 ; w Y w] = [a 0 w Y B n /B n 1 ] = [a 0 ; w Y B n 1 /B n ] = A ny B n 1 /B n + A n 1 B n Y B n 1 /B n + B n 1 = A n 1 + 1n B n Y A n B n There are other forms of symmetry which will appear later so we give the lemma below Note that in all of these cases a 0 = 1 We call these symmetries doubling symmetries following Cohn [6] Lemma 3 26 [1; w Y w] = A n B n n A n B n Y + 1 A n + B n 1 27 [1; w Y w 1] = A n B n 1 n Y A n B n 1

6 6 J MC LAUGHLIN 28 [1; w Y w 1] = A n B n 1 n B n Y A n + 2A n [1; w Y w] = A n 1 n 1 + B n A n B n Y 1 + A n + B n 1 Proof We give the proof only for 26 as and 29 follow similarly We use 24 followed by 21 to get that [1; w Y w] = [1; w Y B n A n B n ] = [1; w Y + 1 A n B n ] = A n Y + 1 An B n + A n 1 B n Y + 1 An B n + B n 1 The result follows from 22 after some simple algebraic manipulation Cohn proved a version of 28 in [6] We also point out that the doubling symmetry described at 26 occurs with some classes of polynomials such as the fx k = 2x + x 2 + x k 1 k xgx mentioned above However S n is not specializable for these polynomials for n k + 1 see Proposition 1 and we have not found S to be specializable for any polynomials that exhibit this kind of doubling symmetry For future reference we show how the various forms of symmetry found in the above lemma will be used Suppose that m when expanded as a continued fraction is equal to S m = [1; w] that the numerator of the ultimate convergent of S m is A m and the denominator of the ultimate convergent is B m and that A m and B m are the numerator and denominator respectively of the penultimate convergent that S m is specializable and that S m+1 is related to S m in one of the ways shown in Lemma 2 or Lemma 3 Y m is used here instead of Y to show the dependence on m Then m+1 = m f m+1 = A m B m On the other hand from the above lemmas 1 + S m+1 = A m B m f m+1 1 HA m B m A m B m Y m where HA m B m A m B m Y m is a polynomial in its variables with integral coefficients that is linear in Y m If solving the equation f m+1 = HA m B m A m B m Y m for Y m leaves Y m in Z[x] for all m then S m is specializable for all m

7 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 7 For later use we also note that if x f + 1 then m simplifies to leave f m in the denominator and say r m in the numerator If f m r m = 1 then up to sign the final numerator convergent of S m is r m and the final denominator convergent is f m A similar situation also holds if x + 1 f As a result of the following lemma polynomials of degree 2 and those of degree 3 or more will be considered separately Lemma 4 If fx has degree greater than 2 then S n+1 contains S n at the beginning of the expansion Proof Suppose S n = [1; a 1 a m ] = p/q where the a i s p and q are polynomials in Q[x] Let [1; a 1 a i ] =: p i /q i and suppose that via the Euclidean algorithm we have that 210 p = q + r 1 q = a 1 r 1 + r 2 r 1 = a 2 r 2 + r 3 r m 2 = a m 1 r m 1 + r m r m 1 = a m r m By definition n+1 = p/q 1 + 1/f n+1 = pf n+1 + 1/qf n+1 and to develop the continued fraction expansion of n+1 one can apply the Euclidean algorithm to this quotient From 210: pf n = q f n+1 + r 1 f n+1 + p qf n+1 = a 1 r 1 f n+1 + p + r 2 f n+1 a 1 p r 1 f n+1 + p = a 2 r 2 f n+1 a 1 p + r 3 f n+1 + p1 + a 1 a 2 Let r 1 = pf n r 0 = qf n+1 and for 1 i m set r i = r i f n i+1 p q i 1 We next show that for 0 i m r i 1 = a i r i + r i+1 This is clearly true for i = 0 1 a 0 = 1 From 210 r i+1 = r i 1 a i r i and from the recurrence relation for the q i s q i+1 = a i+1 q i + q i 1 Suppose 211 is true for i = 0 1 j 1 r j 1 a j r j = r j 1 f n j p q j 2 aj rj f n j+1 p q j 1 = r j 1 a j r j f n j+2 pq j 2 + a j q j 1 = r j+1 f n j+2 p q j = r j+1

8 8 J MC LAUGHLIN Thus 211 is true for 0 i m 1 All that remains to prove the lemma is to show that the degree of r i+1 is less than the degree of r i for 0 i m 1 Let the degree of a polynomial b be denoted by degb From the Euclidean algorithm it follows that degr i+1 < degr i Suppose degf = r 3 so that f i has degree r i and since n i= /f ix = p/q that degp degq 1 + r + r 2 + r n = r n+1 1/r 1 Thus for 0 i m degp q i degp q 2r n+1 1/r 1 < r n+1 = degf n+1 since r 3 This implies that for 0 i m 1 degr i+1 = degr i+1 f n i+2 p q i = degr i+1 f n+1 The result follows < degr i f n+1 = degr i f n i+1 p q i 1 = degr i Note that if degf = 2 so that degf j = 2 j then the situation can be quite different Lemma 5 Let fx be a polynomial of degree two and suppose S n begins with [1; a 1 a k a k+1 ] If k 212 dega k dega i < 2 n+1 then S n+1 begins with [1; a 1 a k ] Proof With the notation of Lemma 4 and its proof [1; a 1 a 2 a k ] will be part of S n+1 if 213 degr i+1 < degr i 0 i k Recall that r i+1 = r i+1f n i+2 p q i so that 213 will follow if degp q i < degr i+1 f n+1 0 i k Let 0 i k Since [1; a 1 a i ] = p i /q i we have that i 214 degq i = dega j j=1 It is clear from 210 that degr j = dega j+1 + degr j+1 This implies that i degr i+1 = degq dega j Now and 215 imply that j=1 degq i + degq degr i+1 < 2 n+1 = degf n+1 The result follows since degp = degq

9 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 9 We return to the case degf 3 The implication of Lemmas 4 and 5 is that if degf 2 then it makes sense to talk of the continued fraction expansion of i= /f i and furthermore that if degf 3 then S is a specializable continued fraction if and only if S n is a specializable continued fraction for each integer n 0 Remark: At this stage we are not concerned with whether the polynomials which are the partial quotients in S have negative leading coefficients or take non-positive values for certain positive integral x Negatives and zeroes are easily removed from regular continued fraction expansions see [21] for example The following lemma means that we get the proof of the specializability of the regular continued fraction expansion of k= /f kx for some classes of polynomials fx for free Lemma 6 Suppose S f is specializable Define gx by 216 gx = f x 1 1 Then S g is specializable Proof If k=0 1+1/f kx has a specializable continued fraction expansion S fx := [1; a 1 x a 2 x ] then k= /f k x 1 has the specializable continued fraction expansion S f x 1 = [1; a 1 x 1 a 2 x 1 ] Let gx be defined as in the statement of the lemma For k 0 g k x = f k x 1 1 This is clearly true for k = 0 1 Suppose it is true for k = 0 1 m Next g m+1 x = gg m x = g f m x 1 1 = f f m x = f m+1 x 1 1 gx = = 1 + gk x = g k x k=0 fk x 1 f k x k=0 k=0 fk x 1 f k x 1 1 From what has been said above the final product has the regular continued fraction expansion [0; 1 a 1 x 1 a 2 x 1 ] and is thus specializable We next demonstrate one of the difficulties in trying to arrive at a complete classification of all polynomials fx for which S f is specializable We need the following lemmas

10 10 J MC LAUGHLIN Lemma 7 Let k be an indeterminate and let t be a non-negative integer Then k t 1 m 2k + k 2 m m=0 where h t k Z[k] = k t+2 h t k + 1 t 2 t+1 1 k t+1 + t 1 m k m m=0 Proof Upon taking the last term on the right side of 217 to the left side and simplifying we get that 1 + k t 1 m 2k + k 2 m m=0 t 1 m k m m=0 = 1 + k 1 [ 2k + k 2 ] t+1 1 [ 2k + k 2 ] = kt+1 [ 2k + k 2 ] t k t kt+1 = k 1 + k 1 kt+1 1 k The final quotient is clearly a polynomial in k with constant term 1 2 t+1 The result now follows Lemma 8 Let k 2 be an integer and let gx Z[x] be such that gx is not the zero polynomial if k = 2 Define fx := 2x + x 2 + x k 1 k + x + 1gx For 0 n k let Then B n = x n j=1 f j f j for some P n x Z[x] f n n = P n x + 2nn 1/2 B n x Proof Since xx + 1 f it follows that B i fi i+1 for i 0 This together with the definition of B n give that fn n = f n n f n + 1 = f n n + f n n 1 B n B n 1 f n B n 1 B n 1

11 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 11 From what has been said just above the first term is in Z[x] and from the definition of fx it follows that for some r n x Z[x] Thus fn n 1 n 1 n 1 fn 1 = r n x + 2 B n 1 B n 1 fn n n 1 n 1 fn 1 = s n x + 2 B n B n 1 for some s n x Z[x] The result follows upon iterating this last expression downwards noting that B 0 = x Proposition 1 Let k 2 be an integer and let gx Z[x] be such that gx is not the zero polynomial if k = 2 Define 219 fx = 2x + x 2 + x k 1 k + x + 1gx Then S n f is specializable for n k If gx 1 k+1 mod x then S n is not specializable for n > k Proof We will show that the doubling symmetry at 26 can be used to develop the continued fraction expansion of n 1 n k More precisely we will show that if S n = [1; w n ] for 0 n k 1 with each partial quotient in S n a polynomial in Z[x] then S n+1 = [1; w n Y n w n ] for some Y n Z[x] We will then show that S k+1 is not specializable unless gx 1 k+1 mod x which would have the effect of replacing k by k +1 in the statement of the form of fx above and this together with Lemma 4 will give the result Note first of all that S 0 = [1; x] and S 1 = [1; x f/xx + 1 x] so that the doubling symmetry at 26 occurs with Y 0 = f/xx + 1 Next let n {0 k 1} and suppose S n = [1; w n ] is specializable We also suppose that S j was developed from S j 1 via the doubling symmetry at 26 for 1 j n so that w n has odd length Let A n /B n denote the final approximant and A n/b n the penultimate approximant of S n We further assume that 220 A n = f n + 1 B n = x n j=1 f j f j Note that this holds for n = 0 1 We also assume that if n 1 then k B n 1 B n j+1 f j n 1 This is true for n = 1 since B 0 = x B 0 = 1 and f 0 = x

12 12 J MC LAUGHLIN By the correspondence between continued fractions and matrices see [22] An A [1; w n ] n B n B n Further from Lemma 1 and its proof An A [1; w n Y n w n ] n Yn 1 Bn B B n B n n 1 0 A n B n B n A n A 2 = n A n B n 1 + Y n B n A n A n A n + A n B n1 + Y n + A nb n B n A n B n Y n B n B n B n A n + B n B n1 + Y n + B n 2 An+1 A =: n+1 B n+1 B n+1 If we set 222 Y n = f n 2 + f 1+n + B n1 + f n B n 1 + f n and use the facts that A n = f n +1 and that w n has odd length so that A n = 1 + B na n /B n = 1 + B nf n + 1/B n by the determinant formula then we get 223 An+1 A n+1 B n+1 B n+1 = 1 + f 1+n 1 + f n B n f 1+n B n B n f 1+n 1 + f n B n 1 f 1+n 1 + f n B n It is clear that A n+1 = 1 + f 1+n1 + f n = 1 + f 1+n = B n+1 f 1+n B n f 1+n n n+1 so that [1; w n Y n w n ] gives the regular continued fraction expansion of n+1 and is specializable provided Y n Z[x] Note also that 220 now holds with n replaced by n + 1 We show Y n Z[x] From the definition of fx we have that f n+1 = 2f n + fn 2 + fn k 1 k f n gf n From 220 and the fact that xx+1 f it follows that B n fn n+1 and since 0 n k 1 B n fn k Thus the result will follow if we can show that 224 B n B n + f n k 1 k 1 or B n B n + 1 j+1 f j n f n + 1 Here and subsequently we mean divisibility in Z[x] We now use the facts clear from 223 that f n B n = B n 1 f n and f n B n = 1 B n 1 f n 1 + 1

13 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 13 to get that 224 will follow if f n B n 1 f n or 225 B n 1 B n 1 B n 1 + k 2 f n f n f n 1 j+1 fn j k 2 1 j+1 fn1 j + f n 1 By the same argument as that just before 224 it follows that B n 1 fn 1 k so that 225 will hold if k B n 1 B n j+1 2f n 1 + fn 1 2 j 1 + f n 1 By 217 k 2 k 2 1 j+1 2f n 1 + fn 1 2 j 1 + f n 1 = 1 j+1 f j n 1 + f k 1 n 1 hf n 1 with hz Z[z] Since B n 1 fn 1 k 1 we can ignore the second term on the right above and increase the index on the sum from k 2 to k 1 for free and get that 226 will hold if k B n 1 B n j+1 f j n 1 However this is true by 221 and thus Y n Z[x] Note that 224 is 221 with n replaced by n + 1 so that the induction can be continued and S n is specializable for 0 n k We next show that if gx = 1 k+1 + b + x g 1 x with b 0 and g 1 x Z[z] then S k+1 is not specializable Define 228 Y k := f k 2 + f 1+k + B k 1 + f k B k 1 + f k + 2kk 1/2 b x Firstly we prove that Y k Z[x] If 219 is used to write f k+1 in terms of f k and we recall that B k f k+1 k it can easily be seen that Y k Z[x] if it can be shown that f k k [ 1 k f k 1 k+1 + b ] + B k 1 + f k B k 1 + f k + 2kk 1/2 b x Z[x]

14 14 J MC LAUGHLIN The first fraction can be re-written as f k+1 k + b k B k By Lemma k+1 + b f k k B k = 1 k+1 + b 1 + fk k /1 + f k + B k B k P n x 2kk 1/2 1 k+1 + b for some P n x Z[x] The second term in 230 can be written as = f k j + B k k 1 k 2 B k = f k j + 1 B k 1 f k 1 + f k 1 k 1 f k j 1 + f k 1 + B k 1 = sx + k 2 B k 1 f B k k 1 1+f k 1 2f k 1 + fk 1 2 j 1 + f k 1 + B k 1 B k 1 for some sx Z[x] Here we have used in turn the formulae from 223 relating B k to B k 1 and B k to B k to write f k in terms of f k 1 and the fact that B k 1 fk 1 k Next we use Lemma 7 to get that 2f k 1 + fk 1 2 j 1 + f k 1 + B k 1 k 2 = B k 1 f k k 1 h k 2f k k 1 2 k 1 1f k 1 = tx + 1 k 1 2 k 1 f k 1 k 1 k 1 B k 1 k 1 k 2 f k 1 j + B k 1 B k 1 k 1 = tx + 1 k 1 f k 1 k B k 1 k 1 x f k 1 j + B k 1 f k 1 j + B k 1 B k 1

15 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 15 for some tx Z[x] Here again we have used the fact that B k 1 f k k 1 Finally Lemma 8 and 221 give that this last expression has the form ux + 1k 1 2 kk 1/2 x for some ux Z[x] Thus 232 f k j + B k k 1 B k = vx + 1k 1 2 kk 1/2 x for some vx Z[x] That Y k Z[x] now follows by and 232 Secondly define α k by Upon solving [1; w k Y k α k] = = A k B k k+1 α k Y k A k + A k + A k α k Y k B k + B k + B k f k+1 = A k B k f k+1 for α k and using 228 to eliminate Y k and the determinant formula to eliminate A k we find B k x 233 α k = 2 kk 1/2 bb k f k B k x Since A k = 1 + f k and k = A k/b k f k and B k have the same degree and same leading coefficient so that 1+f k B k x has degree less than B k x This implies that α k is a rational function whose numerator has higher degree in x than its denominator so that S k+1 begins with [1; w k Y k ] Next x α k 2 kk 1/2 b kk 1/2 b kk 1/2 bb k f k B k x = x B k x f k x + B k x If b = 0 then fx has the form at 219 but with k replaced by k + 1 and from what has been shown already [ ] 1 + f k B k S k+1 = [1; w k Y k w k ] = 1; w k Y k x B k x f k x + B k x The final term in the last continued fraction comes from letting b = 0 on the right side of 234 and is a rational function whose numerator has degree greater than its numerator This must be the case since when b = 0 S k+1 has the form [1; w k Y k x ] as each w k begins with x This implies

16 16 J MC LAUGHLIN that the rational function on the right side of 234 has the same property and so when b 0 [ ] S k+1 = 1; w k Y k x 2 kk 1/2 b + 1 and is thus not specializable The proof is now complete by Lemma 4 Corollary 1 Let k 2 be an integer and let gx Z[x] be such that gx 0 if k = 2 Let fx = x x k k+1 x gx Then S n is specializable for 0 n k If gx 1 k+1 mod x + 1 then S n is not specializable for n > k Proof This follows from Proposition 1 and Lemma 6 One reason we proved Proposition 1 was to show that it is not possible to eliminate all classes of polynomials for which S is not specializable by simply looking at the continued fraction expansion of a finite number of terms of the infinite product for a general polynomial Cohn was able to do this in the infinite series case by looking at just the first four terms 3 Specializability of S for various infinite families of polynomials of degree greater than two We can now show that the specializability of S n occurs for all n for all polynomials in several infinite families We have the following theorem Theorem 1 Let fx Gx and gx denote non-zero polynomials in Z[x] such that the degree of fx is at least three If fx has one of the following forms i fx = x 2 x + 1gx ii fx = xx + 1Gx x 1 iii fx = xx gx 1 iv fx = xx 2 1gx + 2x 2 1 v fx = x + 1xx + 2gx 2x + 1 vi fx = x 2 x 2 1gx + x 2 vii fx = xx + 1x + 2x + 1gx 1 x 2 then for each n 0 S n is a specializable continued fraction Hence S is a specializable continued fraction Proof We note that the proof of iii follows from the proof of i and Lemma 6 and that the proof of v likewise follows from the proof of iv and Lemma 6 However we give independent proofs of iii and v since we also wish to demonstrate the types of doubling symmetry exhibited by

17 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 17 the corresponding continued fractions The proof of vii can similarly be deduced from the proof of vi and in this case no independent proof is given doubling symmetry is not involved for cases vi and vii As in the proof of Proposition 1 throughout let A i /B i denote the final approximant and A i /B i the penultimate approximant of S i = [1; w i ] for each i 0 i For this class of polynomials we will show that S m+1 is derived from S m via the type of symmetry exhibited in the folding lemma Lemma 2 S 0 = [1; x] is clearly specializable Suppose that S m is specializable From Lemma 2 and the discussion following Lemma 3 it is clear that S m+1 is specializable if A m B m f m+1 in Z[x] Since fx = x 2 x + 1gx it follows that for i 0 31 f 2 i f i + 1 f i Since x + 1 f we get after cancellation that = i f i + 1 x i 1 f 2 j gf j Since f j f j+1 for j 0 each term in the denominator of the expression divides f i and thus the numerator and denominator are relatively prime Thus up to sign A i = f i + 1 and B i = f 2 i 1 gf i 1B i 1 The first of these holds for i 0 and the second for i 1 It follows easily by induction that B i f 2 i The facts that B m f 2 m and A m = ±f m +1 together with 31 give that Hence the result A m B m f m+1 ii For this class of polynomial it will be shown that S m is derived from S m 1 by adding a single new partial quotient It is clear from the definition of fx = xx + 1Gx x 1 that for i 0 32 This implies that 33 = i f i + 1 f i+1 f i f i f i f i+2 f i + 1 x i 1 f jgf j 1 x + 1 i 1 = f j + 1Gf j 1 f i This gives that A i f i + 1 and B i f i for all i 0 Next A i+2 = A i f i f i = A i f i+1 + 1Gf i+1 1 B i+2 B i f i+1 f i+2 B i f i+1 Gf i+1 1 We next show that A i f i+1 Gf i+1 1 = B i f i+1 + 1Gf i+1 1 = 1

18 18 J MC LAUGHLIN so that up to sign 34 A i+2 = f i+1 + 1Gf i+1 1A i B i+2 = f i+1 Gf i+1 1B i That B i f i+1 + 1Gf i+1 1 = 1 is easily seen to be true since B i f i f i f i+2 so that B i f i+2 but f i+1 + 1Gf i+1 1 f i+2 +1 The proof that A i f i+1 Gf i+1 1 = 1 is similar We are now ready to prove that S n is specializable for n 0 Initially S 0 = [1; x] and S 1 = [1; x G] It will be shown by induction that S i = [1; α 1 α i+1 ] where all the α j s Z[x] and 1i f i = A i 1 B i Both statements are easily seen to be true for i = 0 1 Suppose these statements are true for i = 0 1 m 1 Let S m 1 = [1; α 1 α m ] Set 35 α m+1 = f m A m 1 Gf m 1 A m 2 which is in Z[x] since A m 1 f m by the remark following 33 Let C m+1 be the numerator of the final convergent of [1; α 1 α m α m+1 ] and let D m+1 be the denominator of the final convergent C m+1 = α m+1 A m 1 + A m 2 = f m 1 + 1Gf m 1 1A m 2 D m+1 = α m+1 B m 1 + B m 2 = f m 1 Gf m 1 1B m 2 The final equality for D m+1 uses the facts that A m 1 B m 2 A m 2 B m 1 = 1 m 1 and 1 m 1 f m 1 = A m 2 B m 1 Hence by 34 C m+1 /D m+1 = A m /B m = m and S m = [1; α 1 α m α m+1 ] Finally A m 1 B m = A m 1 α m+1 B m 1 + B m 2 = f m 1 + 1Gf m 1 A m 2 B m 1 + A m 1 B m 2 = f m 1 + 1Gf m 1 1 m 1 f m m 1 f m m 1 = 1 m f m 1 + 1f m 1 Gf m 1 1 = 1 m f m The third equality also uses the facts that A m 2 B m 1 = 1 m 1 f m 1 and A m 1 B m 2 A m 2 B m 1 = 1 m 1 Hence S n is specializable for all n iii It will be shown that S m+1 is derived from S m via the doubling symmetry found in 27 Suppose S m = [1; w m ] It will be shown that Y m can be chosen such that 36 S m+1 = [1; w m Y m w m 1] Y m Z[x] Note that S 0 = [1; x] and that S 1 = [1; x G x 1] S 1 has even length and if S 2 S m have been defined using 36 then S m has even length It can be seen from 27 that if S m = A m /B m and has even length then f m+1 = A m B m Y m 1 and Y m Z[x] if A m B m f m This we now show

19 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 19 Since fx = xx gx 1 it follows that f j f j After cancellation = x + 1 i 1 f j gf j f i i so that A i x + 1 i 1 f j gf j and B i f i Thus it will be sufficient to show that m 1 f m x + 1 f j gf j f m Suppose that i 1 f i x + 1 f j gf j f i for i = 0 1 m 1 this is clearly true for i = 0 Then m 2 x + 1 f j gf j f m + 1 Since f m gf m 1 f m + 1 it follows that m 1 = f m x + 1 f j gf j f m1 + f m 2 This completes the proof of iii since f m 1 + f m 2 f m iv The argument is similar to that used in the proof of iii It will be shown that S m+1 is derived from S m using the doubling symmetry found in 28 Note that S 1 = [1; x x 1gx 2 x 1] and by induction we assume S m has the symmetric form exhibited in 28 so that A m = B m Note also that the induction means that S m has even length since the duplicating formula always produces a continued fraction of even length It can be seen from 28 that S m+1 = [1; w m Y m w m 1] and will be specializable if the equation 37 B m A m Y m + 2A m = f m is solvable with Y m Z[x] Since fx = xx 2 1gx + 2x 2 1 it can be seen that for i 0 38 f i f i f 2 i 1 f i+1 1

20 20 J MC LAUGHLIN After cancellation 39 m Also 38 implies that = x + 1 m 1 f 2 j 1gf j + 2f j f m m 1 + f j f m 2 1 so that the numerator and denominator in 39 above are relatively prime Thus up to sign B m = f m and A m fm 2 1 Let Y m = f m fm 2 1 gf m B m A m so that Y m Z[x] Upon using the facts that B m = ±f m and from above A m = B m we get that B m A m Y m + 2A m = B m A m Y m + 2B 2 m The result now follows by 37 Cohn also gave a proof of iv in [6] = f m f 2 m 1gf m + 2f 2 m = f m v In this case it will be shown that S m+1 is derived from S m using the doubling symmetry found at 29 Since S 1 = [1; x G x] and w i symmetric implies w i Y i w i is symmetric we have by induction that S m has odd length and that w m is symmetric This gives that B m = A m B m It can thus be seen from 29 that [1; w m Y m w m ] will equal S m+1 and be specializable if the equation 310 f m+1 = A m B m Y m 2 + 2A m leads to Y m Z[x] Since fx = x + 1xx + 2gx 2x + 1 it follows that f j + 1 f j+1 After cancellation 311 = m f m + 1 x m 1 f jf j + 2gf j 2f j + 1 Further since xx + 2 f + 2 it follows that m 1 f j f m + 2

21 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 21 Thus the numerator and denominator in 311 above are relatively prime so that up to sign A m = f m + 1 and B m f m f m + 2 Let Y m = 2 f mf m + 1f m + 2 A m B m gf m so that Y m Z[x] The result now follows from 310 since A m B m Y m 2 2A 2 m = f m + 1f m f m + 2gf m 2f m = f m+1 vi It will be shown that for this class of polynomials and m 1 S m+1 is derived from S m by adding two terms More precisely if m 1 S m = [1; x α 1 β 1 α m β m ] is specializable and 312 α m+1 := f m+1 fm 2 A m B m β m+1 := A m B m then α m+1 β m+1 Z[x] and S m+1 = [1; x α 1 β 1 α m β m α m+1 β m+1 ] Initially S 0 = [1; x] S 1 = [1; x xgxx 1 1 xx + 1] and S 2 = [1; x xgxx 1 1 xx + 1 ff 1gf xx + 1ff + 1] xx + 1 so that 312 holds for m = 1 As part of the proof it will be shown that for i A i = i f j + 1 B i = i f j A i = f i B i 1 These equations are easily shown to be true for i = 1 Suppose that S i has been defined via 312 for i = 2 m that the conditions at 313 are true for i = 1 m and that S m is specializable We first show that α m+1 Z[x] clearly β m+1 Z[x] if S m is specializable Since f = x 2 x 2 1gx+1 we have that x 2 f and x 2 1 f 1 which imply that m 1 m 1 f j f m f j + 1 f m 1 These conditions with 313 imply that A m B m f 2 mf 2 m 1 and hence that A m B m f m+1 f 2 m and thus that α m+1 Z[x]

22 22 J MC LAUGHLIN Since S 0 = [1; x] each S i has odd length in particular S m has odd length Consider the following matrix product: Am A m αm+1 1 βm+1 1 B m B m Am α = m+1 β m A mβ m+1 A m α m+1 + A m B m α m+1 β m B mβ m+1 B m α m+1 + B m A m f m+1 fm A ma m B m = B m f m+1 fm B ma m B m A m f m f m+1 = B m f m B m f m+1 A m Cm+1 C =: m+1 D m+1 D m+1 f m+1 f 2 m B m f m+1 f 2 m A m + A m + B m For the fourth equality we have used the facts induction step that B m = f m B m 1 A m B m A m B m = 1 since S m has odd length and A m = f m /B m 1 By the definition of C m+1 D m+1 C m+1 = A m1 + f m+1 = D m+1 B m f m+1 m f m+1 = m+1 Thus from the relationship between matrices and continued fractions we have that S m+1 = [1; x α 1 β 1 α m β m α m+1 β m+1 ] and Am+1 A A m+1 m f m f m+1 B m+1 B m+1 = B m f m B m f m+1 This equation also implies that 313 holds for i = m + 1 and the result follows A m vii This follows from vi and Lemma 6 4 The Degree Two Case In this section a complete classification is given of all polynomials fx of degree two for which S is specializable or can be transformed in a simple way to produce a continued fraction which is specializable Essentially the method is to start with a general polynomial fx = ax 2 + b 1x + c b 1 a 0

23 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 23 this form makes the continued fraction a little easier to work with and to choose an integer n large enough so that some part of the continued fraction expansion of n say [1; a 1x a t x] forms part of the continued fraction expansion of This follows by Lemma 5 The coefficients in the a i x will be rational functions in a b and c and the requirement that the a i x Z[x] or that S can be transformed to produce a continued fraction that is specializable will impose limiting conditions on a b and c leading to the stated classification Define num := 1 + b + a b c a c 1 + a 2 2 a b + a c + a a b b c x + f 1 + a b + a b a c 1 + a 2 2 a b + a c + a a b 2 x a b a c 1 + a 2 2 a b + a c f 2 den := a b c 1 a a b a c [ 1 + b b 2 + a 2 1 b + c + a 1 + b + b 2 + c b c f a b a 2 + c c b 1 2 a 1 + c + c ] + b c + a b f x [ 1 + b b 2 + a 4 b c + a 1 + b b 2 c + 2 c a b b + c 4 b c + c 2 + a 2 2 b c 2 + b 2 + c b 1 + c ] β := a 1 + a b a c2 1 + a 2 2 a b + a c 2 num a b 4 den Then preferably using a computer algebra system such as Mathematica it can be shown that 41 = [1; 1a + x a 1 b 2 + a c a b 2 a2 x a b 2 a b 2 a 1 + a b a c a 2 2 a b + a c 2 1+b 3 +a 4 b c 2 +a b 4 + 3b + 3c+a 2 1 5b 2 3c 2 + b 3 + 8c + a b 3 2 c + 5 b 2 c + c 3 + b 2 4 c 2 + a b 3 ] x 1 + a b a c 1 + a 2 2 a b + a c β

24 24 J MC LAUGHLIN In what follows we will make use of a remark of Cohn in [6]: that if the first partial quotient in a continued fraction with non-integral coefficients has a non-integral coefficient other than the constant term then the continued fraction is not specializable We will see that some continued fractions with partial quotients in which the constant term is non-integral can be transformed to make them specializable Also polynomials whose coefficients satisfy one of the conditions 42 a b = a b a c = a 2 2 a b + a c = 0 will be considered separately If none of these three equalities hold then the numerator of β has degree four and the denominator has degree three Note that the cofactor of b c + a b f x in den is not zero for any triple of integers a b c This means that if the coefficients of fx do not satisfy one of the conditions at 42 then the next regular partial quotient in S 2 is linear in x so that dega 4 x dega i x = 7 < 2 3 Thus by Lemma 5 S n begins with the first four partial quotients in the continued fraction at 41 if n 2 For specializability it is necessary to have b a a 2 in the third partial quotient the case a = b is to be examined separately Write b a = u 2 v with v square-free Since u 2 a 2 then u a so write a = us Since u 2 v a 2 then v s 2 which implies v s v is square-free or s = v w Thus for specializability it is necessary to have a = u v w b = u 2 v + u v w for some integers u v and w If we substitute for a and b in the coefficient of x in the fourth partial quotient then specializability requires u 6 v u v w c u v u + w 1 + u v w c u v 2 u + w Z A check shows that happens only for or a b c { } f {2x 2 + 2x 2x 2 4x 2 2x 2 1 2x 2 2x 1} That is not specializable for the first and fourth polynomials follows from consideration of S 3 and Lemma 5 We will show that specializability occurs for the third polynomial and specializability for the second will follow from this fact and Lemma 6 We next consider the case a = b proceeding as previously Suppose f = ax 2 + a 1x + c a 1

25 and we define SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 25 num := 1 + a 2 a c [ 1 + f 1 + x 1 + a x 1 + f 3 + f f 1 + x 1 + a x f 1 + x 1 + a x f 3 ] den := a 2 x 1 + x [ a f f 2 f a 1 + c x + a 1 + x + x f 1 + f 3 + f f + f 3 ] β := num den Then preferably once again using a computer algebra system such as Mathematica it can be shown that [ 43 = 1; 1 a + x a + a 3 x 1 a 2 + a c + a 3 x 2 ] 1 a 2 + a c β 3 Further the numerator of β has degree twelve and the denominator has degree ten and the leading coefficient in the numerator or denominator does not vanish except in the case 1 + a 2 a c which is examined separately This all means that apart from this exceptional case the next partial quotient in the regular expansion of 3 has degree two Thus 2 dega 3 x + 2 dega i x = 8 < 2 4 so that S n starts with [ 1; 1 a + x a + a 3 x 1 a 2 + a c + a 3 x 2 ] 1 a 2 + a c for n 3 this once again by Lemma 5 This in turn implies that specializability requires 1 a 2 + a c a 3 and it is not difficult to see that this needs 1 a 2 + a c = ±1 A check shows that the only solutions in this case are or a b c {a a a } f {ax 2 + a 1x 1 x x 2 2x 3 2x 2 + x 2x 2 3x 2} We will show specializability for the case fx = ax 2 + a 1x 1 A more extensive consideration of S 3 shows that S is not specializable for the remaining four of these polynomials Note that for fx = ax 2 +a 1x 1 f x 1 1 = fx so that Lemma 6 gives nothing new We return to the exceptional case 1 a 2 + a c = 0 which is solvable only for a b c { }

26 26 J MC LAUGHLIN or f {x 2 x 2 2x 2} We will show specializability for the first of these polynomials and specializability in the second case will follow from this and Lemma 6 For the exceptional case 1+ab ac = 0 it is clear that a = ±1 is necessary For a = 1 c = b + 1 and an examination of the third partial quotient in S 2 shows b {0 1 2} is necessary Consideration of S 4 eliminates b = 0 and b = 2 using Lemma 5 and b = 1 gives fx = x 2 encountered above For a = 1 c = b 1 and an examination of the third partial quotient in S 2 shows b {0 1 2} is necessary Lemma 5 and consideration of S 4 eliminate b = 0 and b = 2 The case b = 1 gives fx = x 2 2x 2 encountered above Lastly for the exceptional case 1 + a 2 2 a b + a c = 0 it is obvious that a = ±1 is again necessary and in each case c = 2b Consideration of S 3 in the case a = 1 shows that b {0 1 2} is necessary Looking at S 4 eliminates b = 0 and b = 2 and b = 1 gives fx = x 2 which has been encountered above Likewise the case a = 1 necessitates b {0 1 2} Only b = 1 is of interest giving once again fx = x 2 2x 2 The reasoning above leads to the following theorem Theorem 2 Let fx Z[x] be a polynomial of degree two such that f has a specializable continued fraction expansion Then 44 fx {x 2 x 2 2x 2 2 x 2 1 2x 2 4 x 2 a x 2 +a 1x 1} Proof The necessity of 44 has already been shown Also by Lemma 6 it is enough to show sufficiency for the first third and fifth of the polynomials in this list i If fx = x 2 then n 1 + 1fj = i=0 = n i=0 2 n x j x 2n x 2j = x2n +1 1 x 2n x 1 [ ] = 1; x 1 x2n 1 x 1 which is clearly specializable for x 1 and S = [1; x 1]

27 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 27 ii If fx = 2x 2 1 then 45 S 1 = [1; x 1/2 4x 2] S 2 = [1; x 1/2 4x x 4x 2] S 3 = [1; x 1/2 4x x 4x x 4x x 4x 2] We will show that if S n = [1; x 1/2 ω n 4x 2] with ω n specializable then S n+1 = [1; x 1/2 ω n 4x x ω n 4x 2] This can be seen to be true for n = 1 and n = 2 Let T n+1 denote the continued fraction which we claim is equal to S n+1 By induction ω n is made up of the pair of terms 4x x repeated a certain number of times and if T n+1 = S n+1 then it is easy to see that ω n+1 will have the same form We will also show for i 2 that A i = 1 + x2 i+1 i 1 f j and Ai A f i A i i B i B i = 2 A i 4 fi 2 2f 1 i f 46 i A i 2 This is easily checked for i = 2 from 45 Suppose it is true for i = 2 n The continued fraction T n+1 can be constructed as follows: take S n remove the final term 4x 2 add the terms 4x and x and then append another copy of S n which has the first two terms 1 and x 1/2 removed Thus by the correspondence between continued fractions and matrices which we have used several times already An A T n+1 n 0 1 4x 1 x 1 B n B n 1 4x An A n 1 x + 1/2 1 1 B n B n An A = n 1/2 1/2 An A B n B n n 2 0 B n B n A n A n + B n + 4 A n A n A n + 4 A n 2 + A n B n = A n B n + Bn A n B n 2 2 B n A n + B n B n + 4 A n B n A n f n + 2 f n 2 A n f n = f A n + 4 f n + 2 A n f 2 3 n 4 f n n 2 A n =: Cn+1 C n+1 D n+1 D n+1

28 28 J MC LAUGHLIN The next-to-last equality comes from substituting for A n B n and B n from 46 Next C n+1 2 A n f n = D n f = A n 2fn 2 n B n = = fn f n n+1 n+1 so that T n+1 = S n+1 Here we have also used the fact that B n = 2f n It is also now easy to check that 46 now holds with i = n + 1 so that the induction continues Thus S = [1; x 1/2 4x x ] and all that remains is to show that the expansion can be manipulated to remove the 1/2 from the first partial quotient This follows from the identity 47 [x + 1a ] [ ; c α = x; a c + a ] a 2 a2 α If this identity is applied repeatedly it follows that = [1; x 1/2 4x x 4x x 4x x 4x x ] = [1; x 2 x + 1/2 4x x 4x x 4x x 4x ] = [1; x 2 x 2 x 1/2 4x x 4x x 4x x 4x ] = [1; x 2 x 2 ] which is specializable This completes the proof for fx = 2x 2 1 iii If fx = a x 2 + a 1x 1 then 48 S 1 = [1; x 1/a] S 2 = [1; x 1/a a 3 x 2 a 3 x + a] S 3 = [1; x 1/a a 3 x 2 a 3 x + a ax 2 + a 2x 1 + 1/a] S 4 = [1; x 1/a a 3 x 2 a 3 x + a ax 2 + a 2x 1 + 1/a a 3 x 1 + x 1 a x + a 2 x + a 2 x 2 1 a a x + a 2 x + a 2 x 2 ] The situation is somewhat similar to case ii in Theorem 1 going from n to n+1 adds one new term to the continued fraction expansion but the presence of the 1/a term in some partial quotients is troublesome necessitating a different approach Define α 1 α 4 by S 4 = [1; α 1 α 2 α 3 α 4 ]

29 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 29 and for n 2 define 49 n 1 n 1 α 2n+1 = α 3 a f 2i 1af 2i = α 3 f 2i+1 f 2i f 2i + 1f 2i 410 n 1 n 1 f 2i+2 f 2i α 2n+2 = α 4 a f 2i+1 1af 2i = α 4 f 2i+1 + 1f 2i+1 The second equalities follow from the definition of fx It is clear from these definitions and 48 that for n 1 α 2n+2 /a 3 Z[x a] and α 2n+1 1/a Z[x a] We will show that 411 S n = [1; α 1 α n ] for each integer n 1 Let A n /B n denote the final convergent of the right side of 411 As part of the proof we will show that for n A 2n+1 = A 1 1 n A 2n+2 = A 2 1 n B 2n+1 = B 1 1 n B 2n+2 = B 2 1 n n n n n af 2i = A 1 1 n af 2i = A 2 1 n a f 2i 1 = B 1 1 n n a f 2i+1 1 = B 2 1 n n n n f 2i f 2i f 2i+1 f 2i + 1 f 2i+2 f 2i f 2i f 2i+1 Once again the second equalities follow in each case from the form of fx With these values we have for n 1 that A 2n+1 = A n 1 f 2i+1 + 1f 2i + 1 B 2n+1 B 1 f 2i+1 f 2i Similarly = A 1 B 1 = 2n+1 2n+1 i= fi A 2n+2 = B 2n+2 2n+2 for n 1 Thus to prove 411 it is sufficient to prove 412 It is not difficult to check that 412 holds for n = 1 Suppose it holds for n =

30 30 J MC LAUGHLIN 1 2 m A 2m+3 = α 2m+3 A 2m+2 + A 2m+1 = α 3 m = 1 m m f 2i+1 f 2i f 2i + 1f 2i A 2 1 m = 1 m m m+1 = 1 m+1 f 2i f 2i + A 1 1 m m m f 2i f 2i+1 f 2i f 2i f 2m α 3 A 2 + A 1 f f 2i f 2i a A 1 f 2m A 1 f 2i f 2i The next-to-last equality follows from the fact that 413 α 3 A 2 f = a A 1 and the last equality from the fact that f 2m+3 +1 = f 2m+2 af 2m The proof that A 2m+4 has the form stated by 412 is similar except that we use the fact that 414 α 4 A 1 f 2 = a A 2 The proofs that B 2m+3 and B 2m+4 have the forms stated by 412 are similar except that we use in turn the facts that 415 α 3 B 2 = a B 1 f 2 α 4 B 1 f = a B 2 This completes the proof of 411 What remains is to show is that S can be transformed into a specializable continued fraction It is clear from 48 and the remarks following 49 that we can write S = [ 1; x 1 a a3 x 2 + x + a β a a3 β 4 β 2n ] a a3 β 2n+2

31 SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 31 where each β i Z[a x] Proof of specialization now easily from a single application of 47 starting with the first partial quotient S = = = [ 1; x + 1 a a3 x 2 + x + a β a a3 β 4 β 2n ] a a3 β 2n+2 [ 1; x a a3 x 2 + x + a + a a 2 a 2 β a a 3 β 4 a 2 a 2 β 2n ] a3 β 2n+2 a a 2 [ ] 1; x a ax 2 + x a 2 β 3 a aβ 4 a 2 β 2n+1 a aβ 2n+2 which is specializable This completes the proof of Theorem 2 5 Specialization and Transcendence In what follows we assume fx Z[x] and M Z are such that f j M 0 1 for j 0 and f i M f j M for i j For any of the polynomials f in Theorems 1 and 2 S f will typically have some partial quotients which are polynomials in x with negative leading coefficients It may also happen that if S f is specialized by letting x assume integral values that negative or zero partial quotients may appear in the resulting continued fraction These are easily removed as the following equalities show see also [21] [ a b 0 c d ] = [ a b + c d ] [ a b c d e ] = [ a 1 1 b 1 c d e ] Thus if M is an integer repeated application of the identities above will transform S fm to produce the regular continued fraction expansion of the corresponding real numbers A natural question is whether these numbers are transcendental or not We will make use of Roth s Theorem Theorem 3 Roth [14] Let α be an algebraic number and let ɛ > 0 Then the inequality α p q < 1 q 2+ɛ has only finitely many solutions with p Z q N We have the following theorem for the case where the degree of fx is at least three Theorem 4 Let fx Z[x] and M Z be such that f j M 0 1 for j 0 and f i M f j M for i j

32 32 J MC LAUGHLIN If either degf > 3 or degf = 3 and either x f + 1 or x + 1 f then f i M is transcendental i=0 Proof Let f and M satisfy the conditions stated in the theorem and suppose that degf = d and that fx = L x d + a 1 x d a d 1 x + a d =: L x d 1 + βx x Define β i := βf i M so that β i d a i for all i and M Then for k 1 f k M = L f k 1 M d 1 + β k 1 f k 1 M k 1 = L dk 1 d 1 M d k 1 + β i f i M i=0 d k 1 i Note that the second equality for f k M also holds for k = 0 upon taking as usual the empty product to be equal to 1 Also N k=0 Then f k M = L 1 d 1 N k=0 f km d 1 f N+1 M d N+1 1 N+1 d 1 M dn+1 1 d 1 = L N+1 M 1 N N 1 i=0 i=0 1 + β d N i 1 d 1 i f i M 1 + β 1 i f i M Since f i M 0 for any i and the β i are absolutely bounded the product on the right converges so that 51 1 f N+1 M = O 1 N d 1 k=0 km f On the other hand if we set α = fm and p N/q N = N fm in Roth s theorem then it is not difficult to see that α p N = O 1 f N+1 M Since q N N k=0 f km 51 gives that α p N q N = O 1 q d 1 N q N

33 If d 4 then SYMMETRY SPECIALIZABILITY AND INFINITE PRODUCTS 33 α p N < 1 q N q 2+ɛ N has infinitely many solutions for ɛ = 1/2 say and thus fm is transcendental If d = 3 and x f + 1 then q N f N M and since f N+1 M = L f N M β N f N M we get that 52 α p N 1 = O q N so that once again fm is transcendental The case d = 3 and x + 1 f is similar in that in this case p N f N M + 1 Also q N is within a constant factor of p N so that 52 holds and Roth s theorem once more gives transcendence Corollary 2 If fx has any of the forms in the statement of Theorem 1 and M Z is such that f j M 0 1 for j 0 and f i M f j M for i j then fm is transcendental Proof Each polynomial in the statement of Theorem 1 satisfies the conditions of Theorem 4 In the proof of Theorem 4 we were able to show the transcendence of fm when fx had degree three only for the special cases where x f + 1 or x + 1 f If fx Z[x] is a polynomial of degree three such that x f +1 and x+1 f and M is an integer such that f j M 0 1 for any j and f j M f k M for j k is the infinite product f j M transcendental? If this is false find a counter-example With this question in mind we investigated the possibility that a x + b = f j x a x + c for a polynomial fx = r x 3 + s x 2 + t x + u Z[x] and integers a b and c The coefficient of x is the same in the numerator and denominator of the rational function on the right since the infinite product on the left tends to one as x tends to infinity Upon replacing x by fx dividing the new equation into the old and squaring both sides we get x q 3 N 2 a fx + b a fx + c = a x + b a x + c

34 34 J MC LAUGHLIN However comparing coefficients shows that there is no polynomial fx with integral coefficients satisfying 53 Interestingly this approach does lead to the following near miss : if fx = 4x 3 + 6x 2 3/2 and M is any integer different from 1 then M + 3 = f j M 2M 1 It is not evident to the author how to extend Theorem 4 to the remaining polynomials in Z[x] of degree three For the polynomials of degree two in Theorem 2 only fx = a x 2 + a 1x 1 needs investigation We have shown fm converges to a rational number for fx = x 2 M 1 and thus a similar situation holds for fx = x 2 2x 2 by Lemma 6 For fx = 2 x 2 1 fm has an infinite periodic regular continued fraction expansion after removing negatives and zeroes when M 0 ±1 and so fm converges for M 0 ±1 to a quadratic irrational namely signmm + 1/ M 2 1 A similar situation holds for fx = 2 x 2 4 x 2 again by Lemma 6 For fx = a x 2 + a 1x 1 it is not difficult to show from 49 and 412 that if x 1 0 or 1 in the case a = 1 or 2 in the case a = 1 then B 2n+1 54 lim n α 2n+2 can be written as a convergent infinite product If an irrational number α has regular expansion [a 0 ; a 1 ] and its N-th approximant is p N /q N then 55 α p N q N < 1 qn 2 a N+1 for all N 0 If all the negatives are removed from S fm then α 2n+2 will increase or decrease by at most 2 to α 2n+2 say The approximant immediately before α 2n+2 will still be still be A 2N+1/B 2N+1 Thus 54 and 55 will give that fm A 2N+1 = O 1 B 2N+1 3 B 2N+1 and Roth s theorem gives that fm is transcendental We now look at some particular examples of specialization As Cohn showed in [6] if l 2 mod 4 and T k x denotes the k-th Chebyshev polynomial then T l jx has a specializable continued fraction expansion with predictable partial quotients This follows from Theorem 1 iv using the facts that T 1 x =