Some BBP- Functions

Alexandru Lupaş

Our aim is to present a simple proof of the following Proposition 1 If |z + 1| < \sqrt{2} , r ∈ C , then π + 4 arctan z + (2 + 8r) ln \frac{1 − 2z − z^{2}}{1 + z^{2}} = (4 + 8r) ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+1}}{8k + 1} \frac{1}{16^{k}} −

(1) − 8r ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+2}}{8k + 2} \frac{1}{16^{k}} − 4r ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+3}}{8k + 3} \frac{1}{16^{k}} −

− (2 + 8r) ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+4}}{8k + 4} \frac{1}{16^{k}} − (1 + 2r) ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+5}}{8k + 5} \frac{1}{16^{k}} −

− (1 + 2r) ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+6}}{8k + 6} \frac{1}{16^{k}} + r ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+7}}{8k + 7} \frac{1}{16^{k}} .

Choosing r = − \frac{1}{4} , one finds an interesting development of arctan z . More precisely, we obtain Corollary 1 For |z + 1| < \sqrt{2}

π + 4 arctan z = 2 ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+1}}{8k + 1} \frac{1}{16^{k}} + 2 ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+2}}{8k + 2} \frac{1}{16^{k}} +

(2) + ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+3}}{8k + 3} \frac{1}{16^{k}} − \frac{1}{2} ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+5}}{8k + 5} \frac{1}{16^{k}} −

− \frac{1}{2} ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+6}}{8k + 6} \frac{1}{16^{k}} − \frac{1}{4} ∞ ; ∑ ; k=0 \frac{(1 + z)^{8k+7}}{8k + 7} \frac{1}{16^{k}} .

At the same time, using in (2) the equality

arctan 1 + arctan z = arctan t , t := \frac{1 + z}{1 − z} , ( z < 1 ) ,

we give

Corollary 2 If t is real such that |t − 1| < \sqrt{2} , then

arctan t = ∞ ; ∑ ; k=0 \frac{16^{k}}{8k + 1} \frac{t^{8k+1}}{(1 + t)^{8k+1}} + 2 ∞ ; ∑ ; k=0 \frac{16^{k}}{8k + 2} \frac{t^{8k+2}}{(1 + t)^{8k+2}} +

(3) + 2 ∞ ; ∑ ; k=0 \frac{16^{k}}{8k + 3} \frac{t^{8k+3}}{(1 + t)^{8k+3}} − 4 ∞ ; ∑ ; k=0 \frac{16^{k}}{8k + 5} \frac{t^{8k+5}}{(1 + t)^{8k+5}} −

−8 ∞ ; ∑ ; k=0 \frac{16^{k}}{8k + 6} \frac{t^{8k+6}}{(1 + t)^{8k+6}} − 8 ∞ ; ∑ ; k=0 \frac{16^{k}}{8k + 7} \frac{t^{8k+7}}{(1 + t)^{8k+7}} .

For instance , if in (1) we select (z, r) = (0, −\frac{1}{2}) and then let t = 0 in (3) , one finds

π = ∑_{k=0}^{∞} (\frac{4}{8k + 2} + \frac{2}{8k + 3} + \frac{2}{8k + 4} − \frac{1}{2(8k + 7)})\frac{1}{16^{k}} ,

π = ∑_{k=0}^{∞} (\frac{2}{8k + 1} + \frac{2}{8k + 2} + \frac{1}{8k + 3} − \frac{1}{2(8k + 5)} − \frac{1}{2(8k + 6)} − \frac{1}{4(8k + 7)})\frac{1}{16^{k}} .

Let us remark that for (z, r) = (0, 0) we find from (1) the remarkable BBP formula (attributed to David Bailey , Peter Borwein and Simon Plouffe) for π , that is

π = ∞ ; ∑ ; k=0 (\frac{4}{8k + 1} − \frac{2}{8k + 4} − \frac{1}{8k + 5} − \frac{1}{8k + 6})\frac{1}{16^{k}} . (4)

Also when (z, r) = (0, r), r ∈ C one finds another formula for π , namely

π =

= ∞ ; ∑ ; k=0 (\frac{4 + 8r}{8k + 1} − \frac{8r}{8k + 2} − \frac{4r}{8k + 3} − \frac{2 + 8r}{8k + 4} − \frac{1 + 2r}{8k + 5} − \frac{1 + 2r}{8k + 6} + \frac{r}{8k + 7})\frac{1}{16^{k}}

which was discovered by Victor Adamchik and Stan Wagon in their interesting HTML paper [1] . This is a reason why the function L defined as

L(z, r) = π + 4 arctan z + (2 + 8r) ln \frac{1 − 2z − z^{2}}{1 + z^{2}}
( |z + 1| < \sqrt{2} , r ∈ C )

may be called a BBP- function. It is clear that π is expressed as

π = L(0, r)

for any complex r .

Let us denote

(5) σ_{j}(x) = ∞ ; ∑ ; k=0 \frac{x^{8k+j}}{8k + j} , j = 1, 2, ... , |x| < 1 .

According to the fact that for −1 < x < 1 we have

σ_{j}(x) = x ; ∫ ; 0 \frac{t^{j−1}}{1 − t^{8}} dt ,

for j ∈ {1, 2, ..., 7} we find that

8σ_{j}(x) = (−1)^{j+1} ln (1 + x) − ln (1 − x) − π sin \frac{jπ}{4} cos \frac{jπ}{2} −

− ∑_{k=1}^{3} L_{k}(x) cos \frac{kjπ}{4} + 2 ∑_{k=1}^{3} A_{k}(x) sin \frac{kjπ}{4}

(6)

where

L_{k}(x) = ln (1 − 2x cos \frac{kπ}{4} + x^{2}) , A_{k}(x) = arctan \frac{x − cos \frac{kπ}{4}}{sin \frac{kπ}{4}} .

Lemma 1 With the above notation, the following equalities hold on (−1, 1)

\begin{aligned}\sqrt{2}( σ_{1}(x) + σ_{3}(x) − σ_{5}(x) − σ_{7}(x) ) & = & A_{1}(x) + A_{3}(x) \\ 2( σ_{2}(x) − σ_{6}(x) ) & = & A_{1}(x) − A_{3}(x) + \frac{π}{2} \\ 4\sqrt{2}( σ_{1}(x) − σ_{5}(x) ) − 8( σ_{4}(x) + σ_{6}(x) ) & = & E(x) + L(x) \end{aligned}

where the functions E , L : (−1, 1) → R are defined as

(7) E(x) = π + 4 arctan (x\sqrt{2} − 1) , L(x) = 2 ln \frac{1 − x^{2}}{1 − x\sqrt{2} + x^{2}} . According to Lemma 1 we have

\begin{aligned}E(x) & = & 2\sqrt{2}( σ_{1}(x) + σ_{3}(x) − σ_{5}(x) − σ_{7}(x) )+ \\ & & +4( σ_{2}(x) − σ_{6}(x) ) \\ L(x) & = & 2\sqrt{2}( σ_{1}(x) − σ_{3}(x) − σ_{5}(x) + σ_{7}(x) ) − \\ & & −4( σ_{2}(x) + 2σ_{4}(x) + σ_{6}(x) ) \end{aligned} . (8)

Further let us define B : (−1, 1) × C by means of the equality

B(x, r) = E(x) + (1 + 4r)L(x) , (|x| < 1 , r ∈ C) . Lemma 2 The function B has the properties

π = B(\frac{\sqrt{2}}{2}, r) , ∀r ∈ C ,

B(x, r) = 2\sqrt{2}(2 + 4r) ∑_{k=0}^{∞} \frac{x^{8k+1}}{8k + 1} − 16rλ ∑_{k=0}^{∞} \frac{x^{8k+2}}{8k + 2}−

−8\sqrt{2} r ∑_{k=0}^{∞} \frac{x^{8k+3}}{8k + 3} − 8(1 + 4r) ∑_{k=0}^{∞} \frac{x^{8k+4}}{8k + 4} − 4\sqrt{2}(2 + 2r) ∑_{k=0}^{∞} \frac{x^{8k+5}}{8k + 5}−

−8(1 + 2r) ∑_{k=0}^{∞} \frac{x^{8k+6}}{8k + 6} + 8\sqrt{2} r ∑_{k=0}^{∞} \frac{x^{8k+7}}{8k + 7} .

Proof. It is sufficient to use the equalities (5)-(8) and to observe that L(\frac{\sqrt{2}}{2}) = 0 . Now , taking x = \frac{\sqrt{2}}{2}(1 + z) , |1 + z| < \sqrt{2} , then from

L(z, r) = B( \frac{\sqrt{2}(1 + z)}{2}, r )

we conclude with the proof of the desired equality (1).

References

[1] V. Adamchik , S. Wagon , Pi : A 2000 -Year Search Changes Direction, see http : //www.members.wri.com/victor/articles/pi/pi.html

[2] S. Finch , The Miraculous Bailey-Borwein-Plouffe Pi Algorithm, (forthcoming), see http : //www.mathsoft.com/asolve/plouffe/plouffe.html

[3] E. R. Hansen ,A Table of Series and Products, Prentice-Hall, Inc., Englewood Cliffs, N.J., 1975.

Alexandru Lupaş

University ,,Lucian Blaga” of Sibiu Faculty of Sciences Department of Mathematics Str. I. Raţiu nr. 7 2400-SIBIU , Romania E-Mail :lupas@jupiter.sibiu.ro