π in terms of φ via the Machin’s Route

Hei-Chi Chan

Mathematical Science Program, University of Illinois at Springfield

Springfield, IL 62703-5407

email: chan.hei-chi@uis.edu

Abstract

In this paper, we prove some formulas for π that are expressed in terms of

the powers of the reciprocal of the Golden Ratio φ. These formulas depend on

Machin-type identities like the following:

π = 12 arctan(

1φ3

)+ 4 arctan

(1φ5

).

1 Introduction

In a recent paper [18], we proved several formulas of π which are expressed in terms

of the reciprocal of the Golden Ratio φ; for example

π =5√

2 + φ

2 φ

∞∑n=0

(1

2φ

)5n (1

5n + 1+

1

2φ2(5n + 2)− 1

22φ3(5n + 3)− 1

23φ3(5n + 4)

)

(1)

These formulas were inspired by the work of Bailey, Borwein and Plouffe (BBP) [6],

who proved a family of amazing formulas for π with the aid of the powerful PSLQ

1

algorithm [19]. As an example, they proved that

π =∞∑

n=0

1

16n

(4

8n + 1− 2

8n + 4− 1

8n + 5− 1

8n + 6

). (2)

For an introduction and generalizations of (2), see, e.g., [1, 2, 7]; see also the lucid

account in Hijab’s book [20]. For a compendium of currently known results of BBP-

type formulas, see Bailey’s A Compendium of BBP-Type Formulas for Mathematical

Constants, which is available at http://crd.lbl.gov/∼dhbailey.

In this paper, we prove several formulas for π that share similarities with (1): they

express π in terms of the reciprocal of the Golden Ratio φ:

π

4=

∞∑k=0

(−1)k

2k + 1

(1

φ

)2k+1

+

∞∑k=0

(−1)k

2k + 1

(1

φ3

)2k+1

, (3)

= 2∞∑

k=0

(−1)k

2k + 1

(1

φ2

)2k+1

+∞∑

k=0

(−1)k

2k + 1

(1

φ6

)2k+1

, (4)

= 3

∞∑k=0

(−1)k

2k + 1

(1

φ3

)2k+1

+

∞∑k=0

(−1)k

2k + 1

(1

φ5

)2k+1

. (5)

The derivations of these formulas (cf. the next section) are analogous to the formula

discovered by Machin (1680-1752).

Machin’s discovery starts with the following observation:

π

4= 4 arctan

(1

5

)− arctan

(1

239

). (6)

By applying to (6) the power series of arctanx, i.e.,

arctan x =

∞∑k=0

(−1)k x2k+1

2k + 1, (7)

we have the Machin’s formula for π:

π

4=

∞∑k=0

(−1)k

2k + 1

(1

5

)2k+1

−∞∑

k=0

(−1)k

2k + 1

(1

239

)2k+1

. (8)

2

Machin’s discovery plays a key role in computing the digits of π, see [17]. See

also [8, 13]. For generalizations of Machin’s formula, see [15, 17]. See also Weisstein’s

article [22].

It should be note that there are Machin-Type formulas that are closely related to

the Fibonacci numbers. For example, we have

π

4=

∞∑k=1

arctan

(1

F2k+1

).

Cf. chapter 42 of Koshy’s book [21]. See also Ron Knott’s award-winning webstie at

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html.

We also remark that in an interesting paper [14], Jon Borwein, David Borwein

and William Galway explored a class of Machin-type BBP formulas.

Next, we will turn to the proof of (3)-(5).

2 Proofs of the Main Formulas

To prove (3)-(5), all we need to do is to establish the following identities:

π

4= arctan

(1

φ

)+ arctan

(1

φ3

), (9)

= 2 arctan

(1

φ2

)+ arctan

(1

φ6

), (10)

= 3 arctan

(1

φ3

)+ arctan

(1

φ5

). (11)

By using (7), it follows at once that (9) leads to (3), (10) to (4) and (11) to (5). We

will say a few words on how these identities were discovered in the next section.

3

First, we recall two identities that will be useful in our proof: for n ≥ 2

φn = Fnφ + Fn−1 (12)

and for n ≥ 1,

φ−n = (−1)n−1Fnφ + (−1)nFn+1 (13)

Here, Fn is the nth Fibonnaci number. For proofs, see p. 78 of [21], or p. 138 in [12].

The latter is based on probablistic approach; see also the papers [9, 10, 11].

Let us define γ = arctan (1/φ3) . In order to prove (11), we need to show that

tan(π

4− 3γ

)=

1

φ5. (14)

First, we claim that

tan(2γ) =1

2. (15)

Indeed, we have

tan 2γ =2 tan γ

1 − tan2 γ=

2φ−3

1 − φ−6

=2

φ3 − φ−3

=2

F3φ + F2 − (F3φ − F4)=

1

2.

Note that we have used (12) and (13) for φ±3 to derive the third equality.

Next, we show that in a similar manner

tan(3γ) =3 + 2φ

1 + 4φ. (16)

Indeed, by using (15) for tan 2γ we have

tan 3γ =tan γ + tan 2γ

1 − tan γ tan 2γ=

2 + φ3

2φ3 − 1=

3 + 2φ

1 + 4φ.

Note that we have used (12) and (13) to obtain the last equality.

4

With (16), we can establish (14):

tan(π

4− 3γ

)=

1 − tan 3γ

1 + tan 3γ=

φ − 1

3φ + 2=

φ−1

φ4=

1

φ5.

In the third equality, we have used φ − 1 = φ−1 (i.e., n = 1 in (13)) to rewrite the

numerator, and 3φ + 2 = φ4 (i.e., n = 4 in (12)) to rewrite the denominator. This

establishes (11) and implies (5).

The proofs of the other two equations, namely, (9) and (10), follow the same

pattern and we briefly comment on these proofs.

For (9), let us define α = arctan (1/φ) . Then,

tan(π

4− α

)=

1 − tanα

1 + tanα=

φ − 1

φ + 1=

φ−1

φ2=

1

φ3.

Note that we have used the fact that φ− 1 = φ−1 (i.e., n = 1 in (13)) and φ2 = φ + 1

(i.e., n = 2 in (12)). This proves (9).

For (10), let us define β = arctan (1/φ2) . Then, we have

tan 2β =2 tanβ

1 − tan2 β=

2

φ2 − φ−2=

2

2φ − 1.

Note that we have used (12) and (13) to obtain the last line. Finally,

tan(π

4− 2β

)=

1 − tan 2β

1 + tan 2β=

2φ − 3

2φ + 1=

φ−3

φ3=

1

φ6.

This proves (10). �

As a by-product, we note that (15) implies the following identity

arctan

(1

φ3

)=

1

2arctan

(1

2

).

3 Looking Back and Ahead

Originally, identities (9) to (11) were discovered in a series of numerical experiments

using Mathematica. The following is an account of how we discovered (10).

5

Motivated by Machin formula (6), we first set out to look for an identity of the

following form:

π

4= A arctan

(1

φ2

)+ B arctan

(1

φk

). (17)

Our goal was to determine constants A, B and k.

Next, we treated the first term in (17) as a first order approximation of π/4:

π

4� A arctan

(1

φ2

).

We compared this approximation with the following numerical result

π

4� 2.15 arctan

(1

φ2

).

This motivated us to set A = 2 in (17):

π

4= 2 arctan

(1

φ2

)+ B arctan

(1

φk

). (18)

We then looked for B and k that would make (18) an identity. We performed a

series of numerical experiments for this purpose. Precisely, we used Mathematica to

compute, for a range of k, the following ratio which represents B (cf. (18)):

π4− 2 arctan

(1φ2

)

arctan(

1φk

) .

Gladly, we found that at k = 6 (a relatively small k!), Mathematica computed the

ratio to be

1.000000000000000000000000000000

(where we requested the software to give the first 30 decimal places). This suggested

that we should try to prove or disprove the identity

π

4

?= 2 arctan

(1

φ2

)+ arctan

(1

φ6

). (19)

6

With joy and gratitude, we found that (19) turned out to be an exact identity.

One can generalize the present work by considering recursions with three or more

terms. See [22].

We believe that more sophisticated numerical experiments, along the overall theme

suggested in [15, 16], may point to more discoveries of this type of identities. See

also [3, 4, 5].

Acknowledgment. I would like to thank Scott Ebbing for his valuable comments

on this work.

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AMS Subject Classification: 11A05

9