1

MORPHING LORD BROUNCKER’S CONTINUED FRACTION FOR

PI INTO THE PRODUCT OF WALLIS

Thomas J. Osler Mathematics Department

Rowan University Glassboro, NJ 08028

osler@rowan.edu

Introduction

Three of the oldest and most celebrated formulas for pi are:

(1) L21

21

21

21

21

21

21

21

212

+++=π

,

(2) L8897

6675

4453

22312

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

=π

, and

(3)

L++

++

+=

272

52

32

114

2

2

2

2

π.

The first is Vieta’s product of nested radicals from 1592 [6]. The second is Wallis’s

product of rational numbers [7] from 1656 and the third is Lord Brouncker’s continued

fraction [5,7], also from 1656. (In the remainder of the paper we will use the more

convenient notation L+++

+=25

23

2114 222

π for continued fractions.)

In a previous paper [3] the author showed that (1) and (2) are actually special

cases of a more general formula

2

(4)

)(2

122

1221

21

21

21

21

21

212

1

1

11

1

1

radicalskk

kk

kn

n

kn

nn

k+

+∞

=+

+

=

+⋅

−++++= ∏∏ L

π .

By examining this formula for the sequence of special values L,2,1,0=n , we observe

that the product of Wallis (case 0=n ) appears to gradually morph into Vieta’s product as

n approaches infinity. We illustrate this below:

0=n : L12121311

1010119

8897

6675

4453

22312

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

=π

(original Wallis’s product)

1=n : L20202119

16161715

12121311

8897

4453

212

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅=π

2=n : L32323331

24242523

16161715

8897

21

21

21

212

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅+=π

3=n : L48484947

32323331

16161715

21

21

21

21

21

21

21

21

212

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅+++=π

L

∞→n : L21

21

21

21

21

21

21

21

212

+++=π

(original Vieta’s product)

Observe that as we progress through each step of the sequence, one additional factor of

Vieta’s product is added, while every other fraction, starting with the first, in the Wallis

type product is removed.

We will show that Brouncker’s continued fraction (3) and the product of Wallis

(2) are both special cases of the general formula

(5) ⎥⎦

⎤⎢⎣

⎡++++++

+++

= L)14(2

5)14(2

3)14(2

1)14(12

1)(4 222

nnnn

nnW

π,

3

in which

(6) nnnnnW

22)12)(12(

6675

4453

2231)(

⋅+−

⋅⋅

⋅⋅⋅

⋅⋅⋅

= L

is the partial Wallis product. Just as above, this general formula allows us to start with

Lord Brouncker’s continued fraction (case 0=n ) and gradually morph it into the Wallis

product as n approaches infinity.

0=n : L+++

+=25

23

2114 222

π (Lord Brouncker’s continued fraction.)

: 1=n ⎥⎦

⎤⎢⎣

⎡+++

+×⋅⋅

= L105

103

1015

31

22314 222

π

: 2=n ⎥⎦

⎤⎢⎣

⎡+++

+×⋅⋅

⋅⋅⋅

= L185

183

1819

51

4453

22314 222

π

: 3=n ⎥⎦

⎤⎢⎣

⎡+++

+×⋅⋅

⋅⋅⋅

⋅⋅⋅

= L265

263

26113

71

6675

4453

22314 222

π

…

: ∞→n L8897

6675

4453

223124

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

×=π

(original product of Wallis)

Observe that as we progress through each step of the sequence, one additional factor of

Wallis’s product is added, while the Brouncker type continued fraction

⎥⎦

⎤⎢⎣

⎡+++

+ Lxxx

x25

23

21 222

has the value of x incremented by 4.

In a recent paper [2] Lange called attention to a continued fraction for pi

resembling Brouncker’s fraction (3)

4

(7) L+++

+=6

563

613

222

π .

We show that the general formula

(8) ⎥⎦

⎤⎢⎣

⎡++++++

+++

×= L)34(2

5)34(2

3)34(2

1)34()12(

1)(

1 222

nnnn

nnWπ ,

like (5) contains Lange’s continued fraction (7) and the product of Wallis (2) as special

cases. Again by examining this formula as L,2,1,0=n , we can morph (7) into (2) as

shown below:

0=n : L+++

+=6

563

613

222

π . (Lange’s continued fraction.)

1=n : ⎥⎦

⎤⎢⎣

⎡+++

+×⋅⋅

= L145

143

1417

31

3122 222

π .

2=n : ⎥⎦

⎤⎢⎣

⎡+++

+×⋅⋅

⋅⋅⋅

= L225

223

22111

51

5344

3122 222

π .

3=n : ⎥⎦

⎤⎢⎣

⎡+++

+×⋅⋅

⋅⋅⋅

⋅⋅⋅

= L305

303

30115

71

7566

5344

3122 222

π .

…

∞→n : L7566

5344

31222

⋅⋅

⋅⋅⋅

⋅⋅⋅

×=π . (Original product of Wallis reciprocated.)

Extensions of the formulas (6) and (8) are also given.

Derivation of the results and more morphing

All the results of this paper are special cases of the known formula [4, page 35]

5

(9) L+

−+

−+

−+=

⎟⎠⎞

⎜⎝⎛ +−

Γ⎟⎠⎞

⎜⎝⎛ ++

Γ

⎟⎠⎞

⎜⎝⎛ +−

Γ⎟⎠⎞

⎜⎝⎛ ++

Γ

xy

xy

xyx

yxyx

yxyx

25

23

21

41

41

43

434 222222

,

valid for either y an odd integer and x any complex number or y any complex number and

. The names of Euler, Stieltjes, and Ramanujan [1, page 140] have been

associated with this result. Using the very well known formulas

0)Re( >x

!)1( nn =+Γ ,

and )1()( +Γ=Γ xxx π=Γ )2/1( we have πk

kk2

)12(5312

12 −⋅⋅=⎟

⎠⎞

⎜⎝⎛ +

ΓL , valid for

. With this last result and appropriate values of x and y, the left hand side of

(9) can be expressed in terms of rational numbers and

L,3,2,1=k

π . For example, if we set

and in (9) we get our general formula (5) and setting

0=y

14 += nx 0=y and we

get our general formula (8). The manipulations are simple and the reader will have no

difficulty verifying our formulas.

34 += nx

If in (9) we set and ,34 += nx )12(2 += jy for j an integer in the range

we get an extension of (5) nj ≤≤0

(10)

⎥⎦

⎤⎢⎣

⎡++

+−++

+−++

+−++

+−

×−×+++−+−+++−+−

=

L

L

L

)34(2)12(25

)34(2)12(23

)34(2)12(2134

1221

)()222()422)(222()122()322)(122(4

222222222

nj

nj

njn

jn

jnWjnjnjnjnjnjn

π.

Let and (10) becomes 0=j

⎥⎦

⎤⎢⎣

⎡++

−++

−++

−++

+

××++

=

L)34(2

25)34(2

23)34(2

213412

1

)())22()12(4

222222

nnnn

n

nWnn

π .

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We list the special cases of this formula for L,2,1,0=n below:

: 0=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+= L

625

623

6213

214 222222

π

: 1=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+×

⋅⋅

×= L14

2514

2314

21731

2231

434 222222

π

: 2=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+×

⋅⋅

⋅⋅⋅

×= L22

2522

2322

211151

4453

2231

654 222222

π

: 3=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+×

⋅⋅

⋅⋅⋅

⋅⋅⋅

×= L30

2530

2330

211571

6675

4453

2231

874 222222

π

L

: ∞→n L6675

4453

223124

⋅⋅

⋅⋅⋅

⋅⋅⋅

×=π

(original Wallis product)

If in (9) we set and 14 += nx )12(2 += jy for j an integer in the range

we get an extension of (8) nj <≤1

(11) ⎥⎦

⎤⎢⎣

⎡++

+−++

+−++

+−++

−

×−

×+++−+−

++−+−=

L

L

L

)14(2)12(25

)14(2)12(23

)14(2)12(2114

)(21

)(1

)122()5)(322()22()422)(222(

222222222

nj

nj

njn

jn

jnWjnjnjnjnjnjnπ

.

Let and (11) becomes

1=j

⎥⎦

⎤⎢⎣

⎡++

−++

−++

−++

−

×−

×++

+=

L)14(2

65)14(2

63)14(2

6114)1(2

1

)1(1

)32)(12()22)(2(

222222

nnnn

n

nWnnnnπ

.

Let us list this formula for various values on n:

7

: 2=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+×

⋅⋅

×⋅⋅

= L18

6518

6318

61921

3122

7564 222222

π

: 3=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+×

⋅⋅

⋅⋅⋅

×⋅⋅

= L26

6526

6326

611341

5344

3122

9786 222222

π

: 4=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+×

⋅⋅

⋅⋅⋅

⋅⋅⋅

×⋅⋅

= L34

6534

6334

611761

7566

5344

3122

119108 222222

π

: 5=n ⎥⎦

⎤⎢⎣

⎡+

−+

−+×

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

×⋅⋅

= L42

6342

612181

9788

7566

5344

3122

13111210 2222

π

L

: ∞→n L9788

7566

5344

31222

⋅⋅

⋅⋅⋅

⋅⋅⋅

⋅⋅⋅

×=π (original Wallis product reciprocated)

We see that the general formulas (10) and (11) start with a generalized Brouncker type

continued fraction ⎥⎦

⎤⎢⎣

⎡+

−+

−+

−+ L

xy

xy

xyx

25

23

21 222222

and gradually morph it into the

Wallis product as n approaches infinity.

Final remarks

Wallis described the ingenious way in which he obtained his product (2) in [7].

He states that he showed his product to Lord Brouncker who then obtained the continued

fraction (3). It appears that Brouncker never published his method of finding this

continued fraction and only partially explained his reasoning to Wallis. Wallis gives

some hints in [7, pages 167 - 178] as to how Brouncker proceeded but the explanation is

incomplete. Stedall in [5, pages 300-310] has discussed this and made her own conjecture

as to how Brouncker might have reasoned. In his discussion of this question, Wallis

published a table in [7, page 172] which we reproduce here.

8

In the third row of this table we see the continued fractions obtained from our general

formulas (5) and (8). Stedall [5] has recently called attention to these fractions that appear

to have been overlooked. She also points out [5, page 307] that both Wallis and

Brouncker could easily have written the value of these fractions in terms of rational

numbers and pi. Thus we see that the continued fractions that we obtained from (5) and

(8) are among the oldest continued fractions and their values were conjectured as early as

1656!

Acknowledgement

The author wishes to thank James Smoak for his generous assistance with the

historical items in this paper.

References

[1] Berndt, B. C., Ramanujan’s Notebooks, Part II, Springer-Verlag, New York, 1989. [2] Lange, L. J., An Elegant Continued Fraction for π , The American Mathematical Monthly, 106 (1999), pp. 456-458. [3] Osler, T. J., The united Vieta’s and Wallis’s products for pi, American Mathematical Monthly, 106 (1999), pp. 774-776.

9

[4] Perron, O., Die Lehre von den Kettenbruchen, Band II, Teubner, Stuttgart, 1957. [5] Stedall, Jacqueline A., Catching Proteus: The Collaborations of Wallis and Brouncker. I. Squaring the Circle, Notes and Records of the Royal Society of London, Vol. 54, No. 3, (Sep., 2000), pp. 293 -316 [6] Vieta, F., Variorum de Rebus Mathematicis Reponsorum Liber VII, (1593) in: Opera Mathematica, (reprinted) Georg Olms Verlag, Hildesheim, New York, 1970, pp. 398-400 and 436-446. [7] Wallis, John, The Arithmetic of Infinitesimals, (Translated from Latin by Jacqueline A. Stedall), Springer Verlag, New York, 2004.