On the genesis of BBP formulas

TL;DR

This paper introduces a general, elementary method for generating and understanding BBP and BBP-like formulas for transcendental numbers, shedding light on their structure, relations, and origins.

Contribution

It provides a new elementary procedure to produce infinitely many BBP formulas, explaining their interrelations and origins, especially for $ ext{pi}$ and logarithmic constants.

Findings

Derived known BBP formulas for $ ext{pi}$

Explained relations and rearrangements among BBP formulas

Identified sources of null BBP formulas for zero

Abstract

We present a general procedure to generate infinitely many BBP and BBP-like formulas for the simplest transcendental numbers. This provides some insight and a better understanding into their nature. In particular, we can derive the main known BBP formulas for $\pi$. We can understand why many of these formulas are rearrangements of each other. We also understand better where some null BBP formulas representing $0$ come from. We also explain what is the observed relation between some BBP formulas for $\log 2$ and $\pi$, that are obtained by taking real and imaginary parts of a general complex BBP formula. Our methods are elementary, but motivated by transalgebraic considerations, and offer a new way to obtain and to search many new BBP formulas and, conjecturally, to better understand transalgebraic relations between transcendental constants.

Equations

$$\pi =\sum _{k=0}^{+\infty }\frac{1}{16^k}\left(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right).$$

$$\log 2=\sum _{k=1}^{+\infty }\frac{1}{2^k}\frac{1}{k}.$$

$$\pi =\frac{1}{2^6}\sum _{k=0}^{+\infty }\frac{(-1{)}^k}{2^{10k}}\left(-\frac{2^5}{4k+1}-\frac{1}{4k+3}+\frac{2^8}{10k+1}-\frac{2^6}{10k+3}-\frac{2^2}{10k+5}-\frac{2^2}{10k+7}+\frac{1}{10k+9}\right).$$

$$\frac{1}{\pi }=12\sum _{k=0}^{+\infty }\frac{(-1{)}^k(6k)!(545140134k+13591409)}{(3k)!(k!{)}^3(640320{)}^{3k+3/2}}.$$

$$\alpha =Q(d,b,m,\mathbf{A})=\sum _{k=0}^{+\infty }\frac{1}{b^k}\sum _{l=1}^m\frac{a_l}{(km+l{)}^d},$$

$$\alpha =Q(r_0,r_1,d,b,m,\mathbf{A})=r_0+r_1\sum _{k=0}^{+\infty }\frac{1}{b^k}\sum _{l=1}^m\frac{a_l}{(km+l{)}^d},$$

$$I_{n+1}(s)={\int }_1^s{I}_n(u)\mathrm{d}u=\dots ={\int }_1^s{\int }_1^{u_n}\dots {\int }_1^{u_0}\frac{1}{u_0}\mathrm{d}{u}_0\dots \mathrm{d}{u}_{n-1}\mathrm{d}{u}_n.$$

$$I_n(s)=A_n(s)\log s+B_n(s),$$

$$A_n(s)=\frac{s^{n-1}}{(n-1)!},$$

$$I_n(s)=\sum _{j=0}^{+\infty }\frac{(1-s{)}^{j+n}}{n!\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)}=\frac{s^{n-1}}{(n-1)!}\log s+B_n(s),$$

$$\log s=\frac{(1-s{)}^n}{n{s}^{n-1}}\sum _{j=0}^{+\infty }\frac{(1-s{)}^j}{\left(\genfrac{}{}{0ex}{}{j+n}{j}\right)}-\frac{(n-1)!}{s^{n-1}}{B}_n(s).$$

$$\log 2$$

$$\log 2=$$

$$\pi =$$

$$0=\sum _{k=0}^{+\infty }\frac{1}{16^k}\left(\frac{-8}{8k+1}+\frac{8}{8k+2}+\frac{4}{8k+3}+\frac{8}{8k+4}+\frac{2}{8k+5}+\frac{2}{8k+6}-\frac{1}{8k+7}\right).$$

$$P_n(s,t)=\sum _{k=0}^{n-1}\frac{(1-s{)}^k}{k!}{t}^k.$$

$$I_{n+1}(s)={\int }_1^s{I}_n(u)\mathrm{d}u=\dots ={\int }_1^s{\int }_1^{u_n}\dots {\int }_1^{u_0}\frac{1}{u_0}\mathrm{d}{u}_0\dots \mathrm{d}{u}_{n-1}\mathrm{d}{u}_n.$$

$$I_n(s)=(-1{)}^n{\int }_0^{+\infty }\frac{1}{t^n}\left(e^{-st}-P_n(s,t)e^{-t}\right)\mathrm{d}t.$$

$${\int }_0^{+\infty }{e}^{-st}\mathrm{d}t=\frac{1}{s},$$

$$I_{n+1}(s)={\int }_1^s{\int }_0^{+\infty }\frac{1}{t^n}\left(e^{-ut}-P_n(u,t)e^{-t}\right)\mathrm{d}t\mathrm{d}u,$$

$${\int }_1^s{e}^{-ut}\mathrm{d}u$$

$${\int }_1^s{P}_n(u,t)\mathrm{d}u$$

$$e^{-st}-P_n(s,t)e^{-t}=O(t^n).$$

$$I_0(s)={\int }_0^{+\infty }{e}^{-st}\mathrm{d}t=\frac{1}{s}.$$

$$I_1(s)=-{\int }_0^{+\infty }\frac{1}{t}(e^{-st}-e^{-t})\mathrm{d}t=\log s.$$

$$I_1(s)$$

$$I_2(s)$$

$$I_3(s)$$

$$I_4(s)$$

$$I_5(s)$$

$$I_n(s)=A_n(s)\log s+B_n(s),$$

$$A_n(s)=\frac{s^{n-1}}{(n-1)!}.$$

$$B_{n+1}^{\prime }(s)=B_n(s)-\frac{s^{n-1}}{n!}.$$

Full text

1. Introduction

More than 20 years ago, D.H. Bailey, P. Bowein and S. Plouffe ([5]) presented an efficient algorithm to compute deep binary or hexadecimal digits of $\pi$ without the need to compute the previous ones. Their algorithm is based on a series representation for $\pi$ given by a formula discovered by S. Plouffe,

[TABLE]

Formulas of similar form for other transcendental constants were known from long time ago, like the classical formula for $\log 2$, that was known to J. Bernoulli,

[TABLE]

The reader can find in [4] an illustration of the way to extract binary digits from this type of formulae.

Many new formulas of this type, named BBP formulas, have been found for $\pi$ and other higher transcendental constants in the last decades (see [1], [26]). Plouffe’s formula, and others for $\pi$, can be derived using integral periods (as in [5]), or more directly using polylogarithm ladder relations at precise algebraic values (as in [10]), which can be viewed as generalizations of Machin-Störmer relations (see [24] and [25]) for rational values of the arctangent function, and taking its Taylor series expansions. In particular, we can recover in that way Bellard’s formula (see Bellard’s webpage [9]), that seems to be the most efficient one for the purpose of computation of deep binary digits of $\pi$ (see Remark 1.2),

[TABLE]

Many of these formulas are rearrangements of each other, or related by null BBP formulas that represent [math]. The origin of null BBP formulas is somewhat mysterious. Most of the formulas of this sort have been found by extensive computer search over parameter space using the PSLQ algorithm to detect integer relations. So their true origin and nature remainded somewhat mysterious. As the authors of [5] explain:

We found the identity by a combination of inspired guessing and extensive searching using the PSLQ integer relation algorithm.

and in [3]

This formula (1) was found using months of PSLQ computations after corresponding but simpler n-th digit formulas were identified for several other constants, including $\log(2)$. This is likely the first instance in history that a significant new formula for $\pi$ was discovered by a computer.

We note also the observed mysterious numerical relation of BBP formulas for $\pi$ and $\log(2)$.

For the purpose of computation of all digits of $\pi$ up to a certain order, there are more efficient formulas given by rapidly convergent series of a modular nature, initiated by S. Ramanujan ([23]), that are at the origin of Chudnoskys’ algorithm based on Chudnovskys’ formula (see [16])

[TABLE]

Other methods of algorithmic nature include the Borwein quartic algorithm for $\pi$ (see [13]) that approximately quadruples the number of correct digits with each iteration, and the Borwein nonic algorithm for $\pi$ that approximately yields nine-times the number of correct digits.

A general BBP formula as defined in [6] for the constant $\alpha$ is a series of the form

[TABLE]

where $b,d,m$, are integers, $b\geq 2$, and $\mathbf{A}=(a_{1},a_{2},\ldots,a_{m})$ is an integer vector. The integer $d\geq 1$ is the degree of the formula. The classical BBP formula (1) and Bellard formula (2) are of degree $1$. We study in this article formulas of degree $1$. It would be interesting to extend the present results to get higher degree formulas. The integer $b$ is called the base of the BBP formula, and digits in base $b$ can be computed efficiently. Particular attention has been given to base $b=2^{n}$ formulas, as they are useful in computing binary digits. They are called binary BBP formulas. While there are both base $2$ and base $3$ BBP formulas for some constants like $\pi^{2}$ (see [14]), no base $3$ formula for $\pi$ is known.

More generally, we can define BBP-like formulas to be of the general form

[TABLE]

where $r_{0}$ and $r_{1}$ are rational numbers. These more general BBP-like formulas have potentially similar computational applications.

But the interest of these formulas is also theoretical. A normal number in base $b\geq 2$ is an irrational number $\alpha$ such that its expansion in base $b$ contains any string of $n$ consecutive digits with frequency $b^{-n}$. These numbers were introduced in 1909 by É. Borel in an article where he proved that Lebesgue almost every number is normal in any base $b\geq 2$ ([11], and the survey [22]). This result is a direct application of Birkhoff’s Ergodic Theorem to the dynamical system given by the transformation $T:\mathbb{T}\to\mathbb{T}$, multiplication by the base $T(x)=bx$ modulo $1$, where $\mathbb{T}=\mathbb{R}/\mathbb{Z}$. The transformation $T$ preserves the Lebesgue measure which is an ergodic invariant measure. It is not difficult to construct explicit normal numbers, and numbers that are not normal, but there is no known example of “natural” transcendental constant that is normal in every base. It is conjectured that this holds for $\pi$ and other natural transcendental constants, but this remains an open question. It is not even known if a given digit appears infinitely often in the base $10$ expansion for $\pi$. We note recent results [7, 12] where the normality of certain class of constants has been proved, yet not including $\pi$.

An approach to prove normality in base $b$ for any transcendental constant which admits a BBP formula in base $b$ is proposed in [6]. The criterion, named “Hypothesis A”, seems related to Furstenberg’s “multiplication by $2$ and $3$” conjecture (see [17]). Only a very particular class of period-like numbers have BBP formulas (for instante, as mentioned before, $\pi^{2}$ does). It is also natural to investigate the class of numbers with a BBP or BBP-like representation.

The main goal of this article is to present a general procedure to generate the most basic BBP and BBP-like formulas of degree $1$ that correspond to the simplest transcendental numbers $\log p$ and $\pi$. With this new procedure we derive the classical formulas, like Bailey-Borwein-Plouffe or Bellard formulas, and understand better their origin, in particular the origin of null formulas, and the relation of BBP formulas for $\log 2$ and $\pi$ that correspond to take the real or imaginary parts of the same complex formula. We also understand better the redundancy of rearrangements in these formulas, and the method provides a tool to search for more formulas with a more conceptual approach. Although we do not find new BBP formulas, we recover the most important ones and we believe that the method presented can be further developed to discover new ones. We plan to carry this out in the future.

The procedure to generate BBP formulas is elementary and is motivated by considering the bases for first order asymptotics at infinite of Euler Gamma function and higher Barnes Gamma functions and the transalgebraic considerations that play an important role in [20] (see also [19]). To construct these asymptotic bases, we consider the family iterated integrals of $\frac{1}{s}$ defined by $I_{0}(s)=\frac{1}{s}$, and for $n\geq 0$,

[TABLE]

It is elementary to check by induction that

[TABLE]

where $A_{n},B_{n}\in{\mathbb{Q}}[s]$ are polynomials with rational coefficients, with $\deg A_{n}=\deg B_{n}=n-1$, and

[TABLE]

we have

Theorem 1.1**.**

Let $s\in{\mathbb{C}}$, $|s-1|<1$, or $|s-1|=1$ and $n\geq 2$. We have

[TABLE]

or

[TABLE]

Since $B_{n}(s)$ has rational coefficients, we can take $s=1-\frac{1}{b}$ and we get a BBP-like formula for $\log s$. Taking suitable complex values for $s$, and separating real and imaginary parts, we also obtain BBP and BBP-like formulas for $\pi$. We prove that formulas for different values of $n$ provide non-obvious rearrangements of the summations, which in part explains the rich “rearrangement algebra” of BBP formulas.

We recover many formulas with this procedure. For instance, all the formulas of $\log 2$ appearing in Wikipedia [27] are given in (6)–(16). We also get the following classical formulas:

[TABLE]

Also combining our formulas we can get some null formulas representing [math], as for example the following one appearing in [5]

[TABLE]

This gives some explanations of the mysteries mentioned before. For example, formulas for $\log 2$ and $\pi$ are related by taking real and imaginary parts of formulas for complex values for $s$, for example for $s=\frac{1+i}{2}$. Null formulas can appear when comparing our formulas for different complex values of $s$ taking real or imaginary parts. For example for $s=1/2$ and $s=\frac{1+i}{2}$ we do get the previous null formula. It is natural to ask if all null BBP formulas of degree $1$ can be obtaining combining formulas from Theorem 1.1 for different values of $s$.

Certainly, we also recover the classical BBP formula (1) and Bellard formula (2).

Remark 1.2*.*

We can measure the efficiency of a BBP-like formula (3) for computing a number $\alpha$ as ${\bar{m}}/\log b$, where $\bar{m}$ is the number of non-zero coefficients in $\mathbf{A}=(a_{1},a_{2},\ldots,a_{m})$, as this measures the number of non-zero summands for going to the next step in the digit computation. Binary BBP formulas, that is when $b=2$, are of special relevance, since they allow to compute $\alpha$ in binary form. In that case, we can take the logarithm in base $2$. The efficiency of (1) is $1$, whereas the efficiency of (2) is $7/10$, a $43\%$ faster.

The techniques of this article extend to other bases of iterated functions that we will discuss in future articles. We hope that our approach can be useful in finding more efficient BBP-formulas for $\pi$ by more powerful algebraic computer search algorithms.

Acknowledgements

We are very grateful to the anonymous referee that has made a large number of interesting suggestions to improve the exposition. We thank also Tomohiro Yamada for pointing out several corrections. The second author was partially supported by Project MINECO (Spain) PGC2018-095448-B-I00.

2. Laplace-Hadamard regularization of polar parts

The Laplace-Hadamard regularization is related to work in [19] and [20].

For each $n\geq 0$ we define the polynomials $P_{0}=0$, and for $n\geq 1$,

[TABLE]

We also define the iterated primitives of $1/s$ defined by $I_{0}(s)=\frac{1}{s}$, and for $n\geq 0$,

[TABLE]

We call the integrals $I_{n}(s)$ the Laplace-Hadamard regularization or the Laplace-Hadamard transform of $1/t^{n}$. The functions $I_{n}(s)$ are holomorphic functions in ${\mathbb{C}}-]\!-\infty,0]$ and have an isolated singularity at [math] with non-trivial monodromy when $n\geq 1$. We have a single integral expression for $I_{n}(s)$ as a Laplace-Hadamard regularization:

Proposition 2.1**.**

For $n\geq 0$ and $\Re s>0$, or $\Re s=0$ and $n\geq 2$, we have

[TABLE]

Proof.

For $n=0$ we have

[TABLE]

and by induction we get the result integrating on the variable $s$ between $1$ and $s$,

[TABLE]

and using that

[TABLE]

∎

Note that we have $P_{n}(s,t)\to e^{(1-s)t}$ when $n\to+\infty$ uniformly on compact sets, and $P_{n}(s,t)$ is the $n$-th order jet of $e^{(1-s)t}$ at $t=0$. So for $t\to 0$ we have

[TABLE]

For $n=0$ we get the elementary integral

[TABLE]

For $n=1$ we get the old Frullani integral ([12] p.98)

[TABLE]

We have

[TABLE]

A simple induction shows

Proposition 2.2**.**

We have

[TABLE]

where $A_{n},B_{n}\in{\mathbb{Q}}[s]$ are polynomials, with $\deg A_{n}=\deg B_{n}=n-1$, and

[TABLE]

$\Box$**

Regarding the polynomials $B_{n}$, the relation $I^{\prime}_{n+1}(s)=I_{n}(s)$ shows that we have

[TABLE]

This equation with the condition $B_{n+1}(1)=0$ determines $B_{n+1}$ uniquely from $B_{n}$.

We have a formula for $B_{n}$ (see [21], where $I_{n}(s)=f_{n-1}(x)$ with $x=s-1$, and [18]):

Proposition 2.3**.**

We have for $n\geq 0$,

[TABLE]

where $H_{0}=0$ and $H_{n}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$ are the Harmonic numbers.

Proof.

The formula holds for $n=0$ and it satisfies $B_{n+1}(1)=0$ and the recurrence relation:

[TABLE]

∎

Now we prove:

Lemma 2.4**.**

For $n\geq 1$,

[TABLE]

We first establish a useful integral representation for harmonic numbers

Lemma 2.5**.**

[TABLE]

Proof.

We have

[TABLE]

∎

From this it follows

Lemma 2.6**.**

[TABLE]

Proof.

[TABLE]

∎

Now we can prove Lemma 2.4.

Proof of Lemma 2.4..

We have

[TABLE]

∎

Corollary 2.7**.**

For $n\geq 2$,

[TABLE]

Proof.

From (5) we have

[TABLE]

and the result follows from Lemma 2.4. ∎

This is related to the following identity with harmonic numbers:

Lemma 2.8**.**

For $n\geq 2$, we have

[TABLE]

Proof.

For $n\geq 2$, we have

[TABLE]

therefore

[TABLE]

∎

3. Egyptian formulas for rational numbers

We start with the simplest case: an Egyptian formula for rationals. The following is an “infinite Egyptian fraction decomposition” for $\frac{1}{n}$.

Proposition 3.1** (Infinite Egyptian fraction decomposition).**

For $n\geq 2$, we have

[TABLE]

Proof.

Notice that from Proposition 2.1 we have

[TABLE]

with

[TABLE]

hence

[TABLE]

So for $n\geq 2$, we can develop and exchange the integral and the summation:

[TABLE]

Now we have from Lemma 2.4,

[TABLE]

thus

[TABLE]

and the result follows. ∎

As one referee has pointed out to us, Proposition 3.1 follows also by a telescoping sum over

[TABLE]

which is found using Gosper’s algorithm. We show here that this formula results from our general approach.

4. BBP-like formulas for $\log s$

In general we have

Proposition 4.1**.**

For $|s-1|<1$, or $|s-1|=1$ and $n\geq 2$, we have

[TABLE]

Proof.

The condition $\Re s>0$ ensures the convergence of the integrals and $|s-1|<1$, or $|s-1|=1$ and $n\geq 2$ ensures the convergence of the series,

[TABLE]

∎

Remark 4.2*.*

The formula in Proposition 4.1 also holds for $|s-1|=1$ and $s\neq 0$, but the convergence of the sum is only conditional. This can be checked by continuity of both sides making $|s-1|\to 1$.

Now, we have

[TABLE]

and since $B_{n}\in{\mathbb{Q}}[s]$ we get,

[TABLE]

In particular, for $s\in{\mathbb{Q}}$ we have

[TABLE]

Theorem 4.3**.**

Let $|s-1|<1$, or $|s-1|=1$ and $n\geq 2$. Then we have

[TABLE]

∎**

We get a group of formulas for $\log 2$ by specializing at $s=2$. We have

[TABLE]

Using the values $B_{2}(2)=-1$, $B_{3}(2)=-\frac{1}{2}$, $B_{4}(2)=-\frac{8}{9}$, $B_{5}(2)=-\frac{131}{240}$, $B_{6}(2)=-\frac{661}{3600}$, we get:

[TABLE]

Specializing at $s=1/2$, we get the following formula for $n\geq 1$.

[TABLE]

Using the values $B_{2}(1/2)=-2$, $B_{3}(1/2)=-1$, $B_{4}(1/2)=-\frac{40}{36}$, $B_{5}(1/2)=\frac{7}{18}$, $B_{6}(1/2)=-\frac{47}{225}$, we get the formulas:

[TABLE]

All these formulas appear in [27].

It is customary to write the formulas above by splitting the denominators into simple fractions. For instance, the fourth formula can be written as

[TABLE]

If we group for $j=4k,4k+1,4k+2,4k+3$, we get

[TABLE]

We rewrite it in more classical form:

[TABLE]

We can obtain many more binary BBP-like formulas. Specializing at $s=3/2$ we get the formula for $n\geq 1$,

[TABLE]

For instance, $n=4$ gives

[TABLE]

As before, the sum can also be written as

[TABLE]

In general, binary BBP formulas can be obtained from Theorem 4.3 by taking $s=1\pm\frac{1}{2^{N}}$,

[TABLE]

Formulas of this sort are also obtained by Chamberland [15].

The numbers $2$ and $2^{N}\pm 1$, $N\geq 1$, generate a multiplicative subgroup of ${\mathbb{Q}}^{*}$, and for the elements $k$ in that subgroup, we have binary BBP formulas for $\log k$. The first prime that it is not in this subgroup is $k=23$. Note that $2^{11}-1=23\cdot 89$, but these two primes appear always together in the factor decomposition of $2^{N}-1$ when $N$ is a multiple of $11$, and do not appear for other values of $N$. Also they do not appear at all in $2^{N}+1$, for any natural number $N$. This can be checked as follows: first $2^{11}\equiv 1\pmod{23}$, so the order of $2$ in ${\mathbb{Z}}_{23}$ is $11$. In particular it cannot be that $2^{N}\equiv-1\pmod{23}$, since otherwise $2^{2N}\equiv 1\pmod{23}$, and hence $2N|11$, so $N|11$ and thus $2^{N}\equiv 1\pmod{23}$. On the other hand, if $2^{N}\equiv 1\pmod{23}$ then $N$ is a multiple of $11$, and then $23\cdot 89|(2^{11}-1)|(2^{N}-1)$.

5. BBP-like formulas for $\pi$

We may use Theorem 4.3 for a complex value of $s$, then we can get BBP-formulas for $\log k$ and also for $\pi$ separating real and imaginary parts. For $n=1$ (using Remark 4.2), we have

[TABLE]

which is the classical series for $\log s$. Make $s=1+i$. We have $0<\Re(1+i)=1<2$ and

[TABLE]

and

[TABLE]

Separating real and imaginary part and $j=2k$ or $j=2k+1$ we get two BBP formulas, one for $\log 2$ and the other one for $\pi$:

[TABLE]

and

[TABLE]

This last formula is just the first Machin formula for $\pi$, related to

[TABLE]

For general $n\geq 2$, we take $s=1+i$, and we have

[TABLE]

Let

[TABLE]

so that

[TABLE]

With this machinery at hand, we recover a number of known formulas.

Proposition 5.1** (Leibniz).**

We have

[TABLE]

Proof.

We apply the above to $n=2$, where we have that $B_{2}(1+i)=-i$ and $c_{2}=-\Im(-i/(1+i))=1/2$, thus

[TABLE]

∎

The original BBP formula from [5] reads as follows:

Theorem 5.2** (Bailey-Borwein-Plouffe).**

We have

[TABLE]

Proof.

Take $s=\frac{1+i}{2}$, so $\log s=-\frac{1}{2}\log 2+i\frac{\pi}{4}$. Using the formula for $n=1$, we have

[TABLE]

Taking the imaginary part, and agroupping terms for $j=8k+r$, $r=1,2,\ldots,7,8$, we get

[TABLE]

so

[TABLE]

Similarly, by taking the real part, we get

[TABLE]

so

[TABLE]

Substracting (19) and our previous formula (17), we get a null formula

[TABLE]

(note that the term $k=0$ gives exactly $1/105=71/105-2/3$). Adding (18) to twice this formula, we get

[TABLE]

∎

In the proof we have proved and used the following null BBP formula that appears in [5] :

Proposition 5.3**.**

We have

[TABLE]

Null BBP formulas are very interesting and useful for rewritting BBP formulas. They are obtained by comparing BBP formulas for the same number at different values of $s$.

Proposition 5.4**.**

We have

[TABLE]

Proof.

We use the formulas

[TABLE]

Adding the first two and substracting the third, we get

[TABLE]

and multiplying by $32$ we get the result. ∎

Finally, we include a proof of Bellard’s formula.

Theorem 5.5** (F. Bellard).**

We have

[TABLE]

Proof.

We use the following factorization

[TABLE]

and taking imaginary parts

[TABLE]

For $s=(7+i)/8$ and $n=1$, we get

[TABLE]

writing $j=8k+r$, $r=1,2,\ldots,8$, and then $2k=l$.

Now take $s=1+i/2$ and $n=1$, to get

[TABLE]

We substract twice (23) minus (22), and use that $2\frac{8}{10l+5}-\frac{8}{4l+2}=-\frac{4}{10l+5}$. Then we get the result. ∎

6. On the classical BBP form

As defined in [6] the classical BBP form is

[TABLE]

where $b,d,m$ are integers and $A=(a_{1},a_{2},\ldots,a_{m})$ is a vector of integers. The degree is $d$ and the base is $b$. Let us check that with our formula from Theorem 4.3 we get BBP formulas of degree $1$.

Lemma 6.1**.**

We have

[TABLE]

where for $l=1,2,\ldots,n$, $c_{l}$ is an integer given by

[TABLE]

Proof.

As usual, multiply by $j+l$ and set $j=-l$ to get

[TABLE]

∎

We have a general reorganization Lemma that shows that any sum of BBP form with more than $m$ fractions can be reorganized into one with $m$ terms.

Lemma 6.2**.**

We have

[TABLE]

with

[TABLE]

where the sum extends over indexes $l+1\leq i\leq n$ such that $mj+i=mk+l$.

Proof.

For $k=0,1,\ldots$ and $l=1,\ldots m$ group the fractions of the sum modulo $m$ with $mj+i=mk+l$. ∎

These two Lemma prove that the BBP formulas that we get from Theorem 1.1 are of type $Q(1,b,1,(a_{1}))$.

We can apply this reorganization to the summation in the formula from Theorem 1.1 and get (regrouping the terms with $j=k-l+1$ in the third equality),

[TABLE]

with $a_{k}=\sum\limits_{l=1}^{n}c_{l}(1-s)^{n-l}$. But we have

[TABLE]

Hence, we recognize in the last sum of (6) $\log s$, so the formula in Theorem 1.1 for $n\geq 2$ is a rearrangement of the formula for $n=1$ that is the classical Taylor formula for $\log s$

[TABLE]

We can use this rearrangement to recover directly the formula for the polynomials $B_{n}$ directly:

[TABLE]

but

[TABLE]

and

[TABLE]

which gives after some rearrangment the expression for $B_{n}(s)$

Formally, there is no extra content in the formulas for the same parameter $s$ but different integers $n\geq 2$. However, these rearrangements are computationally useful, and they are not easy to produce. The iterated integrals $I_{n}(s)$ or Proposition 2.2 gives a systematic method to find a family of such resummations. The expression in terms of combinatorical coefficients in the denominator that arise by the iterated integrals in this type of sums can present sometimes some advantages. Of course one is inmediately reminded (even if it is a formula of higher degree) of the famous Apery sum for $\zeta(3)$ starting point of his proof of the irrationality of this number.

Appendix. Location of the zeros of the polynomials $B_{n}$

The application of the formula in Theorem 4.3 to roots of $B_{n}$, in particular to real roots, gives BBP-like formulas of a special form. We study the location of the roots of $B_{n}$ and the number of real roots.

To understand the polynomials $B_{n}(s)$, we introduce the polynomials $C_{n}(x)$ of degree $n-2$, for $n\geq 2$, defined by

[TABLE]

so that by Proposition 2.3

[TABLE]

We list the polynomials:

[TABLE]

and accordingly,

[TABLE]

We want to locate the zeros of $C_{n}(x)$.

Lemma 6.3**.**

We have

[TABLE]

Proof.

The value at $x=-1$ follows from Lemma 2.4. The value at $x=0$ by (27). ∎

Let

[TABLE]

The zeros of $D_{n}$ are those of $C_{n}$ and an extra zero at $x=0$. Now we have two interesting equalities:

[TABLE]

and

[TABLE]

Using these equalities, we can prove the following:

Proposition 6.4**.**

For $n\geq 2$ even, the polynomial $C_{n}(x)$ has no real roots.

For $n\geq 3$ odd, the polynomial $C_{n}(x)$ has only one real root and it lies in the interval $]\!-1,0[$.

Proof.

We want to prove by induction that:

• •

For $n$ even, $x=0$ is the only (simple) zero of $D_{n}(x)$. And $D_{n}(x)<0$ for $x<0$ and $D_{n}(x)>0$ for $x>0$.

• •

For $n$ odd, $D_{n}$ has two zeros, at some $x_{0}\in]\!-1,0[$ and at $x=0$. And $D_{n}(x)>0$ for $x\in]\!-\infty,x_{0}[\,\,\cup\,\,]0,\infty[$ and $D_{n}(x)<0$ for $x\in]x_{0},0[$.

Let $n$ be even. We want to prove that $D_{n}(x)$ has only a zero at $x=0$. Note that $D_{n}(0)=0$ and $D_{n}^{\prime}(0)=1$, so $D_{n}$ is increasing at $x=0$. For $n$ even we have $Q_{n}(x)>0$ everywhere.

• •

If $x\leq-1$ then $D_{n-1}(x)<0$ by induction hypothesis. By (Appendix. Location of the zeros of the polynomials $B_{n}$) we have $D_{n}^{\prime}(x)>0$, so it is increasing there. By Lemma 6.3, $D_{n}(-1)<0$ so there are no zeros on $]\!-\infty,-1]$.

• •

If $x>0$ then $D_{n-1}(x)>0$ by induction hypothesis. By (Appendix. Location of the zeros of the polynomials $B_{n}$) we have $D_{n}^{\prime}(x)>0$, so it is increasing there. As $D_{n}(0)=0$, there are no zeros on $]0,\infty[$.

• •

If $x\in]\!-1,0]$ then $Q_{n}(x)>0$. If $D_{n}(x)=0$ then (Appendix. Location of the zeros of the polynomials $B_{n}$) says that $(1+x)D_{n}^{\prime}(x)>0$. So $D_{n}$ is increasing at every zero. As $x=0$ is a zero, then this implies that there is only one zero of $D_{n}$.

Now let $n$ be odd. We want to prove that $D_{n}(x)$ has a zero at some $x_{0}\in]\!-1,0[$ and at $x=0$, it is positive on $]\!-\infty,x_{0}[\,\,\cup\,\,]0,\infty[$ and negative at $]x_{0},0[$. Note that $D_{n}(0)=0$ and $D_{n}^{\prime}(0)=1$, so it is increasing at $x=0$. Note that for $n$ odd we have $Q_{n}(x)>0$ for $x>-1/2$, and $Q_{n}(x)<0$ for $x<-1/2$.

• •

If $x\leq-1$ then $D_{n-1}(x)<0$ by induction hypothesis. By (Appendix. Location of the zeros of the polynomials $B_{n}$) we have $D_{n}^{\prime}(x)<0$, so it is decreasing there. By Lemma 6.3 $D_{n}(-1)>0$, so there are no zeros on $]\!-\infty,-1]$.

• •

If $x>0$ then $D_{n-1}(x)>0$ by induction hypothesis. By (Appendix. Location of the zeros of the polynomials $B_{n}$) we have $D_{n}^{\prime}(x)>0$, so it is increasing there. As $D_{n}(0)=0$, there are no zeros on $]0,\infty[$.

• •

If $x\in]\!-1/2,0[$ then (Appendix. Location of the zeros of the polynomials $B_{n}$) says that $(1+x)D_{n}^{\prime}(x)>(n-1)D_{n}(x)$. So if there is a zero, $D_{n}$ is increasing. As the last zero before $x=0$ cannot be increasing, this last zero has to be $x_{0}\leq-1/2$.

• •

For $x=-1/2$, if it was a zero of $D_{n}$, then it is also a zero of $D_{n}^{\prime}$ because of (Appendix. Location of the zeros of the polynomials $B_{n}$). Then we write $x=-1/2+h$, and develop (Appendix. Location of the zeros of the polynomials $B_{n}$) to see that $D_{n}^{\prime}(-1/2+h)>0$ for $h>0$ small. But this implies that there must be another zero of $D_{n}$ in $]\!-1/2,0[$ with decreasing slope, which contradicts the previous item.

• •

If $x\in]\!-1,-1/2[$ then (Appendix. Location of the zeros of the polynomials $B_{n}$) says that $(1+x)D_{n}^{\prime}(x)<(n-1)D_{n}(x)$. So if there is a zero, $D_{n}$ is decreasing. There must be at least one zero, but there cannot be two zeros, since there cannot be two decreasing consecutive zeros.

∎

It is relevant to locate the complex zeros of $C_{n}(x)$. The polynomial $C_{4}$ has a pair of conjugate complex roots $x\approx-0.68182\pm 0.28386i$. The polynomial $C_{5}$ has one real root $x_{0}\approx-0.61852$ and a pair of conjugate complex roots $x\approx-0.73074\pm 0.49200i$. The polynomial $C_{6}$ has $2$ pairs of conjugate complex roots: $x\approx-0.18252\pm 0.39103i$, $x\approx-1.2226\pm 0.9258i$. We may expect that all roots of $C_{n}(x)$ have $\Re x\in]\!-\infty,0[$.

To locate the complex roots of $C_{n}(x)$, we rewrite the differential equation (Appendix. Location of the zeros of the polynomials $B_{n}$) as

[TABLE]

Take $f_{n}(x)=(1+x)^{-(n-1)}D_{n}(x)$, hence $\mathrm{d}f_{n}=(1+x)^{-n}\left((1+x)^{n-1}-x^{n-1}\right)\mathrm{d}x$. We make the change of variables $w=\frac{x}{1+x}$ to get $\mathrm{d}f_{n}=\frac{1-w^{n-1}}{1-w}\mathrm{d}w=(1+w+....+w^{n-2})\mathrm{d}w$, and integrating

[TABLE]

where we have used that for $w=0$, it is $x=0$ and hence $f_{n}=0$. Note that $f_{n}(w)$ is the truncation of the series $-\log(1-w)$, which is convergent on $|w|<1$.

Proposition 6.5**.**

The polynomial $f_{n}(w)$ has no roots in $|w|\leq 1$ except $w=0$.

Proof.

We will look at the polynomial

[TABLE]

for which we want to check that the only root in the disc $|w|\leq 1$ is $w=1$. For $|w|\leq 1$, we have

[TABLE]

Then if $Q(w)=0$, we have

[TABLE]

which implies $|w-2|\leq 1$. Combined with $|w|\leq 1$, we have $w=1$. ∎

Undoing the change of variables $w=\frac{x}{1+x}$, we get that all roots of $C_{n}(x)$ are in $\Re x<-\frac{1}{2}$. Therefore, with (26) we get that the roots of $B_{n}(s)$ are $s=1$ and the others lie in $\Re s<\frac{1}{2}$.