Infinite products involving binary digit sums

Infinite Products Involving Binary Digit Sums

arXiv:1709.04104v1 [math.NT] 13 Sep 2017

Samin Riasat

1 Introduction Throughout this article n denotes a non-negative integer. Let sb (n) denote the sum of the digits in the base-b expansion of n. Although we will always take b = 2, all our results can be easily extended to other values of b. Put un = (−1)s2 (n) . In other words, un is equal to 1 if the binary expansion of n has an even number of 1’s, and is equal to −1 otherwise. We study infinite products of the form n + b un . f (b, c) := ∏ n≥1 n + c (We will show in Sect. 2 that f (b, c) converges for any b, c ∈ C \ {−1, −2, −3, . . . }.) It seems that the only known non-trivial value of f (up to the relations f (b, b) = 1 and f (b, c) = 1/ f (c, b)) is √ 1 f , 1 = 2, 2 which is the famous Woods-Robbins identity [8, 9]. Several infinite products inspired by this were discovered afterwards (see, e.g., [5, 7]). But none of them involve the sequence un . In this article we will compute another value of f , namely, 1 3 3 f , = . 4 4 2 In Sect. 2 we study some general properties of f and introduce a related function h. In Sect. 3 we prove some analytical results on h. In Sect. 4 we try to find infinite

Samin Riasat University of Waterloo, Waterloo, Ontario, Canada, e-mail: [email protected]

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Samin Riasat

products of the form ∏ R(n)un admitting a closed form value, with R a rational function.

2 General Properties of f and a New Function h First we establish a general result on convergence from [3]. Lemma 1. Let R ∈ C(x) be a rational function such that the values R(n) are defined and non-zero for n ≥ 1. Then, the infinite product ∏n R(n)un converges if and only if the numerator and the denominator of R have same degree and same leading coefficient. Proof. If the infinite product converges, then R(n) must tend to 1 when n tends to infinity. Thus the numerator and the denominator of R have the same degree and the same leading coefficient. Now suppose that the numerator and the denominator of R have the same leading coefficient and the same degree. Decomposing them in factors of degree 1, it suffices, for proving that the infinite product converges, to show that infinite products of the form n + b un ∏ n≥1 n + c converge for complex numbers b and c such that n + b and n + c do not vanish for any n ≥ 1. Since the general factor of such a product tends to 1, it is equivalent, grouping the factors pairwise, to prove that the product 2n + b u2n 2n + 1 + b u2n+1 ∏ 2n + c 2n + 1 + c n≥1 converges. Since u2n = un and u2n+1 = −un we only need to prove that the infinite product (2n + b)(2n + 1 + c) un ∏ n≥1 (2n + c)(2n + 1 + b) converges. Taking the (principal determination of the) logarithm, we see that (2n + b)(2n + 1 + c) 1 =O 2 log (2n + c)(2n + 1 + b) n which gives the convergence result. t u Hence f (b, c) converges for any b, c ∈ C\{−1, −2, −3, . . . }. Using the definition of un we see that f satisfies the following properties.


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Properties of f For any b, c, d ∈ C \ {−1, −2, −3, . . . }, 1. f (b, b) = 1. 2. f (b, c) f (c, d) = f(b, d). b c c+1 c+1 b+1 f , , 3. f (b, c) = f . b+1 2 2 2 2

One can ask the natural question: is f the unique function satisfying these properties? What if we impose some continuity/analyticity conditions? Properties 1 and 2 above give f (b, c) f (d, e) =

f (b, c) f (c, d) f (d, e) f (d, c) f (b, e) f (d, c) = = f (b, e) f (d, c). f (c, d) f (d, c) f (c, c)

Hence we can rewrite property 3 as c c+1 b b+1 f , , f 2 2 2 2 . f (b, c) = b+1 c+1 Thus f (b, c) can be computed using only the quantities h(b) = f ( 2b , b+1 2 ) via f (b, c) =

c + 1 h(b) · . b + 1 h(c)

(1)

So understanding f is equivalent to understanding h, in the sense that each can be completely evaluated in terms of the other. Moreover, taking c = b + 12 in Eq. (1) gives the functional equation b+1 1 h(b) = h b+ h(2b). (2) 2 b + 32 An approximate plot of h is given in Fig. 1. Similar questions can be asked for h: is it the unique solution to Eq. (2)? What about monotonic/continuous/smooth solutions? A New Function The function h(x) := f ( 2x , x+1 2 ) satisfies the functional equation x+1 1 h(x) = h x+ h(2x). 2 x + 32

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Fig. 1 An approximate plot of h(x)

3 Analytical Properties of h Recall that

∞

h(x) = ∏

n=1

2n + x 2n + 1 + x

un .

Lemma 2. For b, c ∈ (−1, ∞), 1. if b = c, then f (b, c) = 1. 2. if b > c, then

c+1 b+1

2 < f (b, c) < 1.

3. if b < c, then 1 < f (b, c)
c > −1 and put an = log

n+b , n+c

N

SN =

∑ an un ,

N

UN =

n=1

∑ un .

(3)

n=1

Note that an is positive and strictly decreasing to 0. Using t2n + t2n+1 = 1, it follows that Un ∈ {−2, −1, 0} and Un ≡ n (mod 2), for each n. Using summation by parts, N

SN = aN+1UN + ∑ Un (an − an+1 ). n=1


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So −2a1 < SN < 0 for large N. Exponentiating and taking N → ∞ gives the desired result. t u Lemma 2 together with Eq. (1) implies the following results. Theorem 1. h(x)/(x +1) is strictly decreasing on (−1, ∞) and h(x)(x +1) is strictly increasing on (−1, ∞). Theorem 2. For b, c ∈ (−1, ∞), f (b, c) is strictly decreasing in b and strictly increasing in c. Theorem 3. For x ∈ (−2, ∞), 1 < h(x)
M. Thus Sn0 converges uniformly on (−2, ∞), which shows that log h, hence h, is differentiable on (−2, ∞). Now suppose that derivatives of h up to order k exist for some k ≥ 1. Note that (k+1) SN

N k

= (−1) k! ∑ un n=1

As before,

1 1 − (2n + x)k+1 (2n + 1 + x)k+1

.

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(k+1) (k+1) − SM ≤ k! SN

N

1 1 − (2n + x)k+1 (2n + 1 + x)k+1

1 1 − (2n − 1 + x)k+1 (2n + 1 + x)k+1

∑

n=M+1 N

≤ k!

∑

n=M+1

k! k! − k+1 (2M + 1 + x) (2N + 1 + x)k+1 k! < →0 (2M − 1)k+1 =

(k+1)

as M → ∞, for any x ∈ (−2, ∞) and N > M. Hence Sn converges uniformly on (−2, ∞), i.e., h(k) is differentiable on (−2, ∞). Therefore, by induction, h has derivatives of all orders on (−2, ∞). t u Theorem 5. Let a ≥ 0. Then (−1)k−1 log h(x) = log h(a) + ∑ k k=1 ∞

∞

un ∑ (n + a)k n=2

! (x − a)k

for x ∈ [a − 1, a + 1]. Proof. Let H(x) = log h(x). By Theorem 4, ∞

un . k+1 n=2 (n + x)

H (k+1) (x) = (−1)k k! ∑ Hence

∞

∞ 1 1 ≤ k! ∑ k+1 |n + x| (n + a − 1)k+1 n=2 n=2

|H (k+1) (x)| ≤ k! ∑

for x ∈ [a − 1, a + 1]. So by Taylor’s inequality, the remainder for the Taylor polynomial for H(x) of degree k is absolutely bounded above by ! ∞ 1 1 ∑ (n + a − 1)k+1 |x − a|k+1 k + 1 n=2 which tends to 0 as k → ∞, since a ≥ 0 and |x − a| ≤ 1. Therefore H(x) equals its Taylor expansion about a for x in the given range. t u

4 Infinite Products Recall that

f (b, c) = ∏

n≥1

n+b n+c

un .


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Properties 1–3 in Sect. 2 give

∏

n≥1

c (n + b)(n + b+1 2 )(n + 2 )

(n + c)(n +

!un =

c+1 b 2 )(n + 2 )

c+1 b+1

for any b, c 6= −1, −2, −3, . . . , and if b, c 6= 0, −1, −2, . . . , then !un c (n + b)(n + b+1 2 )(n + 2 ) = 1. ∏ b c+1 n≥0 (n + c)(n + 2 )(n + 2 )

(4)

(5)

Some interesting identities can be obtained from Eqs. (4)–(5). For example, in Eq. (4), taking c = (b + 1)/2 gives !un (n + b)(n + b+1 b+3 4 ) (6) ∏ (n + b+3 )(n + b ) = 2(b + 1) n≥1 4 2 while taking b = 0 gives

∏

n≥1

(n + 21 )(n + 2c ) (n + c)(n + c+1 2 )

!un = c+1

(7)

for any b, c 6= −1, −2, −3, . . . . We now turn our attention to the functional equation Eq. (2); recall that b+1 1 h(b) = h b + h(2b). 2 b + 32 Taking b = 0 gives 1 2 h(0). h(0) = h 3 2 Since 1 < h(0) < 9/4 by Theorem 3, cancelling h(0) from both sides gives h(1/2) = 3/2. This shows that 4n + 3 un = 2. (8) ∏ n≥0 4n + 1 Next, taking b = 1/2 in Eq. (2) gives 3 1 h = h(1)2 2 4 √ hence h(1) = 2 (since 1 < h(1) < 16/9 by Theorem 3) and we recover the WoodsRobbins product 2n + 2 un √ = 2. (9) ∏ n≥0 2n + 1

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Samin Riasat

Similarly, taking b = −1/2 in Eq. (2) gives 1 1 1 1 1 1 1 1 h − = h(0)h(−1) = f 0, f − ,0 = f − , , 2 2 2 2 2 2 2 2 i.e.,

∏

n≥1

(4n − 1)(2n + 1) (4n + 1)(2n − 1)

un

1 = . 2

(10)

Taking b = 1 in Eq. (2) gives 4 3 h(1) = h h(2) 5 2 √ hence h(3/2)h(2) = 5 2/4 and this gives 1 (4n + 3)(2n + 2) un ∏ (4n + 5)(2n + 3) = √2 . n≥0

(11)

Taking b = 3/2 in Eq. (2) and using the previous result gives 3 h(2)2 h(3) = √ 2 which is equivalent to

∏

n≥0

(2n + 2)(n + 1) (2n + 3)(n + 2)

un

1 =√ . 2

(12)

Eqs. (8)–(12) can also be combined in pairs to obtain other identities.

5 Concluding Remarks The quantity h(0) ≈ 1.62816 appears to be of interest [1, 4]. It is not known whether its value is irrational or transcendental. We give the following explanation as to why h(0) might behave specially in a sense. Note that the only way non-trivial cancellation occurs in the functional equation Eq. (2) is when b = 0. Likewise, non-trivial cancellation occurs in Eq. (2) or property 3 in Sect. 2 only for (b, c) = (0, 1/2) and (1/2, 0). That is, the victim of any such cancellation is always h(0) or h(0)−1 . So one must look for other ways to compute or study h(0). √ Using the only two known values h(1/2) = 3/2 and h(1) = 2, the following expressions for h(0) are obtained from Theorem 5: • by taking x = 0 and a = 1,


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∞ √ un 1 ∞ h(0) = 2 exp − ∑ ∑ k k=1 k n=2 (n + 1)

! .

• by taking x = 1 and a = 0, √ h(0) = 2 exp

(−1)k ∞ un ∑ k ∑ nk n=2 k=1 ∞

! .

• by taking x = 0 and a = 1/2, ∞

1 ∞ u2n+1 ∑ k ∑ (2n + 1)k k=1 n=2

3 h(0) = exp 2

! .

• by taking x = 1/2 and a = 0, 3 h(0) = exp 2 The Dirichlet series

∞

un

(−1)k ∞ u2n ∑ k ∑ (2n)k n=2 k=1

∑ (n + 1)k

n=0

∞

∞

and

! .

un

∑ nk

n=1

appearing in the above expressions were studied by Allouche and Cohen in [2]. Acknowledgements This work is part of a larger joint work [3] with Professors Jean-Paul Allouche and Jeffrey Shallit. I thank the professors for helpful discussions and comments.

References 1. J.-P. Allouche, Thue, Combinatorics on words, and conjectures inspired by the Thue-Morse sequence, J. de Th´eorie des Nombres de Bordeaux, 27, no. 2 (2015), 375–388. 2. J.-P. Allouche, H. Cohen, Dirichlet series and curious infinite products, Bull. London Math. Soc. 17 (1985), 531–538. 3. J.-P. Allouche, S. Riasat, J. Shallit, More infinite products: Thue-Morse and the Gamma function, preprint. 4. J.-P. Allouche, J. Shallit, The ubiquitous Prouhet-Thue-Morse sequence. Sequences and their Applications: Proceedings of SETA ’98, (1999), 1–16. 5. J.-P. Allouche, J. Shallit, Infinite products associated with counting blocks in binary strings, J. London Math. Soc. 39 (1989), 193–204. 6. J.-P. Allouche, J. Shallit, Automatic Sequences. Theory, Applications, Generalizations, Cambridge University Press, Cambridge, 2003. 7. J.-P. Allouche, J. Sondow, Infinite products with strongly B-multiplicative exponents, Ann. Univ. Sci. Budapest. Sect. Comput. 28 (2008), 35–53. [Errata: Ann. Univ. Sci. Budapest. Sect. Comput. 32 (2010), 253.] 8. D. Robbins, Solution to problem E 2692, Amer. Math. Monthly 86 (1979), 394–395. 9. D. R. Woods, Elementary problem proposal E 2692, Amer. Math. Monthly 85 (1978), 48.