# Topics in Probability Theory and Stochastic Processes Steven R

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Steven R. DunbarDepartment of Mathematics203 Avery HallUniversity of Nebraska-LincolnLincoln, NE 68588-0130http://www.math.unl.edu

Voice: 402-472-3731Fax: 402-472-8466

Topics in

Probability Theory and Stochastic ProcessesSteven R. Dunbar

Wallis’ Formula

Rating

Mathematically Mature: may contain mathematics beyond calculus withproofs.

1

Section Starter Question

Can you think of a sequence or a process that approximates π? What is theintuition or reasoning behind that sequence?

Key Concepts

1. Wallis’ Formula is the amazing limit

limn→∞

(2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n− 1) · (2n+ 1)

)=π

2.

2. One proof of Wallis’ formula uses a recursion formula developed fromintegration of trigonometric functions.

3. Another proof uses only basic algebra, the Pythagorean Theorem, andthe formula πr2 for the area of a circle of radius r.

4. Yet another proof uses Euler’s infinite product representation for thesine function.

Vocabulary

1. Wallis’ Formula is the amazing limit

limn→∞

(2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n− 1) · (2n+ 1)

)=π

2.

2

Mathematical Ideas

Introduction

Wallis’ Formula is the amazing limit

limn→∞

(2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n− 1) · (2n+ 1)

)=π

2. (1)

Another way to write this is

π

2=∞∏

j=1

(2j)2

(2j − 1)(2j + 1).

A “closed form” expression for the product in Wallis formula is

limn→∞

24n · (n!)4

((2n)!)2(2n+ 1)=π

2

or equivalently

limn→∞

24n

(2nn

)2(2n+ 1)

=π

2.

Note that Wallis Formula is equivalent to saying that the “central bino-mial term” has the asymptotic expression

1

22n

(2n

n

)∼√

2

(2n+ 1)π.

See the subsection Central Binomial below for a proof of an equivalent in-equality.

In the form

wn =n∏

j=1

(2j)2

(2j − 1)(2j + 1)=

24n

(2nn

)2(2n+ 1)

(2)

3

0

0.5

1

1.5

2wn

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20n

π/2

Figure 1: Convergence of the Wallis formula to π/2.

4

it is easy to see that the sequence wn is increasing since 4n2/(4n2 − 1) > 1.Figure 1 illustrates the increasing sequence.

Doing a numerical linear regression of log(π/2− wn) versus log n on thedomain n = 1 to n = 30 indicates that wn approaches π/2 at a rate which isO(1/n).

A proof using integration and recursion

Theorem 1 (Wallis’ Formula).

limn→∞

(2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n+ 1)

)=π

2. (3)

Proof. Consider Jn =∫ π/20

cosn(x) dx. Integrating by parts with u = cosn−1(x)and dv = cos(x) shows

∫ π/2

0

cosn(x) dx = (n− 1)

∫ π/2

0

cosn−2(x) sin2(x) dx

= (n− 1)

∫ π/2

0

cosn−2(x)(1− cos2(x)) dx

= (n− 1)

∫ π/2

0

cosn−2(x) dx−(n− 1)

∫ π/2

0

cosn(x) dx .

Gathering terms, we get nJn = (n− 1)Jn−2.Now J1 = 1 so recursively J3 = 2

3, J5 = 2·4

3·5 and inductively

J2n+1 =2 · 4 · · · (2n− 2) · (2n)

1 · 3 · · · (2n− 1) · (2n+ 1). (4)

Likewise J2 = π2·2 , and J4 = 3·π

2·4·2 , J6 = 3·5·π2·4·6·2 and inductively

J2n =3 · 5 · · · (2n− 3) · (2n− 1) · π

2 · 4 · · · (2n− 2) · (2n) · 2 .

For 0 ≤ x ≤ π/2, 0 ≤ cos(x) ≤ 1, so cos2n(x) ≥ cos2n+1(x) ≥ cos2n+2(x),implying in turn that J2n ≥ J2n+1 ≥ J2n+2. Then

1 ≥ J2n+1

J2n≥ J2n+2

J2n=

2n+ 1

2n+ 2.

5

Hence limn→∞J2n+1

J2n= 1.

That is,

limn→∞

(2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n+ 1)

2

π

)= 1

or equivalently

limn→∞

(2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n+ 1)

)=π

2.

An elementary proof of Wallis formula

The following proof is adapted and expanded from [8]. The proof uses onlybasic algebra, the Pythagorean Theorem, and the formula πr2 for the areaof a circle of radius r. Another important property used implicitly is thecompleteness property of the reals.

Define a sequence of numbers by s1 = 1 and for n ≥ 2,

sn =3

2· 5

4· · · 2n− 1

2n− 2. (5)

These are the reciprocals of the subsequence J2n−1 defined in equation (4).The partial products of Wallis’ formula (1) with an odd number of terms

in the numerator are

on =22 · 42 · · · (2n− 2)2 · 2n

1 · 32 · · · (2n− 1)2=

2n

s2n, (6)

while those with an even number of factors in the numerator are of the form

en =22 · 42 · · · (2n− 2)2

1 · 32 · · · (2n− 3)2 · (2n− 1)=

2n− 1

s2n. (7)

Interpret e1 = 1 as an empty product. Since (2n)2

(2n−1)(2n+1)> 1, clearly en <

en+1. Since (2n)(2n+2)(2n+1)2

< 1, clearly on > on+1. Also, en < on. Therefore

e1 < e2 < e3 < · · · en < on < · · · < o3 < o2 < o1

6

for any n. Furthermore, for 1 ≤ i ≤ n,

2i

s2i= oi ≥ on

and2i− 1

s2i= ei ≤ en

from which it follows that

2i− 1

en≤ s2i ≤

2i

on. (8)

For convenience, define s0 = 0 so that inequality (8) holds also for i = 0.Denote the successive difference an = sn+1 − sn so a0 = s1 − s0 = 1 and forn ≥ 1

an = sn+1 − sn = sn

(2n+ 1

2n− 1

)=sn2n

=1

2· 3

4· · · 2n− 1

2n.

Lemma 2. For the sequence ai we have the identity

aiaj =j + 1

i+ j + 1aiaj+1 +

i+ 1

i+ j + 1ai+1aj (9)

for any i and j.

Proof. Make the substitutions

ai+1 =2i+ 1

2(i+ 1)ai

and

aj+1 =2j + 1

2(j + 1)aj

the right side of the proposed identity (9) becomes

aiaj

(2j + 1

2(j + 1)· j + 1

i+ j + 1+

2i+ 1

2(i+ 1)· i+ 1

i+ j + 1

)= aiaj.

7

Lemma 3.

1 = a20 = a0a1 + a1a0

= a0a2 + a21 + a2a0

. . .

= a0an + a1an−1 + · · ·+ ana0

Proof. Start from a20 = 1 and repeatedly apply the identity (9). At stage napplying the identity (9) to every term the sum

a0an + a1an−1 + · · ·+ ana0

becomes(a0an +

1

na1an−1

)+

(n− 1

na1an−1 +

2

na2an−2

)+ · · ·+

(1

nan−1a1 + ana0

).

Collecting terms, this simplifies to a0an + a1an−1 + · · ·+ ana0.

Now divide the positive quadrant of the xy-plane into rectangles by draw-ing the vertical lines y = sn and the horizontal lines y = sn for all n. LetRi,j be the rectangle with lower left corner (si, sj) and upper right corner(si+1, sj+1). The area of Ri,j is aiaj. Thus the identity 1 = a0an + a1an1 +· · ·+ ana0 states that the total area of the rectangles Ri,j for which i+ j = nis 1. Let Pn be the polygonal region consisting of all rectangles Ri,j for whichi+ j < n. Hence the area of Pn is n.

The outer corners of Pn are the points (si, sj) for which i+ j = n+ 1 and1 ≤ i, j,≤ n. By the Pythagorean theorem, the distance of such a point to

the origin is√s2i + s2j . By (8)

√2(i+ j)

on=

√2(n+ 1)

on.

bounds this distance from above. Similarly the inner corners of Pn are thepoints (si, sj) for which i + j = n and 0 ≤ i, j ≤ n. The distance of such apoint to the origin is bounded from below by

√2(i+ j − 1)

en=

√2(n− 1)

en.

8

Therefore, Pn contains a quarter circle of radius√

2(n− 1)/en and is con-

tained in a quarter circle of radius√

2(n+ 1)/on. See Figure 2 for a diagramof the polygonal region Pn and the corresponding inner and outer quartercircles for n = 4.

The area of a quarter circle of radius r is πr2 and the area of Pn is n.This leads to the bounds

(n− 1)π

2en< n <

(n+ 1)π

2on(10)

from which it follows that

(n− 1)π

2n< en < on <

(n+ 1)π

2n.

Then as n→∞, en and on both approach π/2.

Wallis’ Formula and the Central Binomial Coefficient

This subsection gives a detailed proof that Wallis’ Formula gives an ex-plicit inequality bound on the central binomial term that in turn impliesthe asymptotic formula for the central binomial coefficient. The derivationin [1]. motivates this proof.

Start from the expansion (2) for the Central Binomial Coefficient:

wn =n∏

j=1

(2j)2

(2j − 1)(2j + 1)=

24n

(2nn

)2(2n+ 1)

.

Rearrange it to

(2n

n

)2

=16n

2n+ 1

n∏

j=1

(2j − 1)(2j + 1)

(2j)2

and use the definitions (6) and (7) of the partial products of the Wallisformula to obtain (

2n

n

)2

=16n

(2n+ 1)

2n+ 2

(2n+ 1)on+1

and (2n

n

)2

=16n

(2n+ 1)

1

en+1

.

9

y

xs1

s1

s2

s2

s3

s3

s4

s4

r =√

2(4 + 1)/o4r =√2(4− 1)/e4

R0,0

R0,1

R0,2

R0,3

R1,0

R1,1

R1,2

R2,0

R2,1

R3,0

Figure 2: A diagram of the regions Ri,j and the inner and outer quartercircles for the case n = 4.

10

Rearrange the inequality (10) to obtain

2n+ 2

π(n+ 2)<

1

on+1

and1

en+1

<2n+ 2

nπ.

Then

16n

(2n+ 1)

(2n+ 2)2

(2n+ 1)(n+ 2)π<

(2n

n

)2

<16n

(2n+ 1)

(2n+ 2)

nπ

and taking square roots and slightly rearranging again

4n√(2n+ 1)(π/2)

√2(n+ 1)2

(2n+ 1)(n+ 2)<

(2n

n

)<

4n√(2n+ 1)(π/2)

√(n+ 1)

n.

To simplify, take a series expansion of the square roots of the rational ex-pressions and truncate, leaving

4n√(2n+ 1)(π/2)

(1− 1

2n

)<

(2n

n

)<

4n√(2n+ 1)(π/2)

(1 +

1

2n

).

A proof using the product expansion of the sine function

Theorem 4. The expansion of sin(z) as an infinite product is

sin(z) = z∞∏

m=1

(1− z2

m2π2

).

Proof. See [7, page 312] for the proof. The article in the Mathworld.comarticle on http://mathworld.wolfram.com/Sine.html also gives references toEdwards 2001, pages 18 and 47; and Borwein et al. 2004, page 5.

Although the common proof uses complex analysis, as in the texts citedabove, a proof using only elementary analysis is possible. The following proofis adapted from [4].

Proof. Start with the definition of the Chebyshev polynomials Tn(x) fromthe trigonometric identity cos(nx) = Tn(cos(x)). Then

cos(2kx) = Tk(cos(2x)) = Tk(1− 2 sin2 x).

11

Together with the sine product identity

sin((2k + 1)x)− sin((2k − 1)x) = 2 sin(x) · cos(2kx)

this inductively shows that there is a polynomial Fm of degree m such that

sin((2m+ 1)x) = sin(x) · Fm(sin2(x)).

(In fact, using the definition Un(cos(x)) = sin((n+1)x)sin(x)

for the Chebyshev poly-

nomials of the second kind, it is easy to show that Fm(x) = U2n(√

1− x).)Substituting xk = kπ

2m+1and noting that sin((2m+ 1)xk) = 0 shows that Fm

has zeros at sin2(( kπ2m+1

)for k = 1, 2, . . . ,m. These zeros are distinct, so Fm

has no other zeros, then

Fm(y) = Fm(0)m∏

k=1

(1− y

sin2(

kπ2m+1

))

and

Fm(0) = limx→0

sin((2m+ 1)x)

sin(x)= 2m+ 1.

Therefore

sin((2m+ 1)x) = (2m+ 1) sin(x)m∏

k=1

(1− sin2(x)

sin2(

kπ2m+1

))

and changing variables

sin(x) = (2m+ 1) sin

(x

2m+ 1

) m∏

k=1

(1−

sin2(

x2m+1

)

sin2(

kπ2m+1

)). (11)

The goal now is to estimate the product terms. For all real t, et ≥ 1+t andtherefore e−t ≤ 1

1+t. For u < 1, the choice t = u/(1−u) gives e−u/(1−u) ≤ 1−u.

Then for every collection of numbers uk ∈ [0, 1), we have

1−∑

k

uk1− uk

≤ e−

∑k

uk1−uk ≤

∏

k

(1− uk) ≤ e−∑

k uk ≤ 1. (12)

If also∑

k uk < 1, then we also know that

e−∑

k uk ≤ 1

1 +∑

k uk(13)

12

and subtracting the first and third terms of (12) from 1 and using (13)

∑k uk

1 +∑

k uk≤ 1−

∏

k

(1− uk) ≤∑

k

uk1− uk

. (14)

Let m and N be positive integers with m > N . Take x such that |x| <14Nπ and x

π/∈ Z. Then define uk by

uk =

(sin(

x2m+1

)

sin(

kπ2m+1

))2

, k = 1, 2, . . . ,m.

Use (11) by dividing the leading factor and the first N factors onto the leftside to obtain

sin(x)

(2m+ 1) sin(

x2m+1

)∏Nk=1(1− uk)

=m∏

k=N+1

(1− uk).

Then use the first, third and fifth terms of (12) to see that

1−m∑

k=N+1

uk1− uk

≤ sin(x)

(2m+ 1) sin(

x2m+1

)∏Nk=1(1− uk)

≤ 1. (15)

For 0 ≤ t ≤ π2

we have sin(t) ≥ 2πt, so we see that

uk ≤(

(2m+ 1) sin(

x2m+1

)

2k

)2

≤( x

2k

)2;

thusuk

1− uk≤ x2

(2k)2 − x2 fork > N.

Hencem∑

k=N+1

uk1− uk

≤ x2

2N − |x| .

Thus it follows from (15) that

1− x2

2N − |x| ≤sin(x)

(2m+ 1) sin(

x2m+1

)∏Nk=1(1− uk)

≤ 1.

13

Let m→∞ so that

1− x2

2N − |x| ≤sin(x)

x∏N

k=1(1− uk)≤ 1.

Now let N →∞ and we obtain for xπ/∈ Z

sinx = x

∞∏

k=1

(1− x2

k2π2

).

For xπ∈ Z this equality is also true.

The following somewhat probabilistic proof of Euler’s infinite productformula is adapted from [2].

Proof. Start with the integral∫ n

0

xt−1(

1− x

n

)ndx

and integrate by parts once:

u =(

1− x

n

)nv =

1

txt

du = −(

1− x

n

)n−1dx dv = xt−1 dx .

Then∫ n

0

xt−1(

1− x

n

)ndx =

(1− x

n

)n∣∣∣n

0+

∫ n

0

1

txt(

1− x

n

)n−1dx

=1

t

∫ n

0

xt(

1− x

n

)n−1dx

Integrate by parts again:

u =(

1− x

n

)n−1v =

1

t+ 1xt+1

du = −n− 1

n

(1− x

n

)n−2dx dv = xt dx

14

so ∫ n

0

xt−1(

1− x

n

)ndx =

(n− 1)

t(t+ 1)n

∫ n

0

xt+1(

1− x

n

)n−2dx .

After integration by parts n times:∫ n

0

xt−1(

1− x

n

)ndx =

(n− 1)!

t(t+ 1) . . . (t+ n− 1)nn−1

∫ n

0

xt+n−1 dx

=(n− 1)!

t(t+ 1) . . . (t+ n− 1)nn−1xt+n

t+ n

∣∣∣∣n

0

=n!nt

t(t+ 1) . . . (t+ n).

Then by dominated convergence it follows that

Γ(t) =

∫ ∞

0

xt−1e−x dx = limn→∞

n!nt

t(t+ 1) . . . (t+ n).

Note that limit definition of the Gamma function is also http://dlmf.nist.gov/5.8E1. It follows that for −1 < t < 1 that

Γ(1 + t)Γ(1− t) =∞∏

k=1

1

1− t2/k2 .

For X and Y independent exponential random variables, both with ex-pectation 1 we have that

E[X t]

=

∫ ∞

0

xte−x dx = Γ(1 + t)

and likewise E [Y −t] = Γ(1− t). Then using the independence and symmetry

Γ(1 + t)Γ(1− t) = E[X t]E[Y −t

]= E

[(X/Y )t

]= E

[(X/Y )|t|

]

The distribution function for the random variable X/Y is

P [X/Y ≤ u] = P [Y ≥ X/u] = E[e−X/u

]=

u

u+ 1

for u > 0. Then the probability density is

d

duP [X/Y ≤ u] =

1

(1 + u)2, u > 0.

15

This gives by integration by parts, for −1 < t < 1, that

E[(X/Y )|t|

]=

∫ ∞

0

u|t|

(u+ 1)2du =

∫ ∞

0

|t|u|t|−1u+ 1

du

The last integral is πt/ sin(πt). For example, this is formula 613, page445 in the 18th Edition of the CRC Standard Mathematical Tables. It canalso be done with contour integration in the complex plane

∫ +∞

0

|t|u|t|−1u+ 1

du−2πi(−1)|t|−1 +

∫ 0

+∞

|t|u2πi(|t|−1)u+ 1

du = 0

so ∫ ∞

0

|t|u|t|−1u+ 1

du(1− e2πi|t|

)= −2πieπi|t|.

Thus ∫ ∞

0

|t|u|t|−1u+ 1

du =|t|2πieπi|t|

e2πi|t| − 1=

π|t|sin(π|t|) =

πt

sin(πt).

Hence, for −1 < t < 1

Γ(1 + t)Γ(1− t) = E[(X/Y )t

]=

πt

sin(πt).

A standard integration-by-parts gives the fundamental Gamma function re-cursion Γ(t) = Γ(1 + t)/t. Using this to define Γ(t) for t < 0, it follows thatfor all real t,

sin(πt) =π

Γ(t)Γ(1− t) .

Combining all results above,

sin(πt) = πt∞∏

k=1

(1− t2

k2

).

Corollary 1 (Wallis’ Formula).

limn→∞

2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n− 1) · (2n+ 1)=π

2.

Proof. Substitute z = π/2 in the product expansion of sin(z).

Remark. In [5] Ciaurri uses Tannery’s Theorem for Infinite Products, a trigidentity for cotangent and tangent, and Wallis formula to provide an ele-mentary proof of product expansion of the sine function. In this sense, theproduct expansion of the sine function is equivalent to Wallis’ formula.

16

A Geometric Interpretation of Wallis’ Formula

This section gives a geometric interpretation of Wallis’ formula, constructinga subset of the unit square with area 8

9· 2425· 4849· · · = 2

3· 43· 45· 65· 67· 87· · · = 1

2· π2.

The construction is reminiscent of the construction of the Sierpinski Triangle.Take a unit square W0 and remove a square of area 1

3· 13

from the center.Label the result as W1 with area 8

9. Next remove a square of area 1

5· 15

fromthe center of each of the 8 small sub-squares of area 1

9constituting W1 and

call the result W2 with area 89· 2425

. Continuing by removing squares of area17· 17

from each sub-square of area 15· 15

surrounding the hole removed at stage2, construct region W3 with area 8

9· 2425· 4849

. Continue to create W4, and soon. Call the limiting region W∞ which has area 8

9· 2425· 4849· · · = 1

2· π2.

The side length of the removed squares at stage n is 2nn!(2n+1)!

. After 3stages, 201 squares have been removed, and the stage 3 holes have side length13· 15· 17

= 1105

. It will be impractical to illustrate more than 3 stages of thisiterative process.

Figure 3: Three stages in the construction of the Wallis sieve.

Here area means the Lebesgue measure of the limiting region W∞. How-ever W∞ does not contain any product set A×B with A,B ⊂ R, each withpositive Lebesgue measure. To see this, consider a maximal product subsetof W1, for example [0, 1]×([0, 1

3]∪ [2

3, 1]) (or equivalently ([0, 1

3]∪ [2

3, 1])× [0, 1]

although from here on, consider only the product sets with first factor [0, 1]).This product subset has measure 2

3. A similar maximal product subset of W2

has measure 23· 45. Continuing, a maximal product subset of W∞ has measure

23· 45· 67· · · =

∞∏n=1

(1− 1

2n+1

)= 0.

Thus, W∞ is an example of a subset of the plane R2 with positive Lebesguemeasure, but not admitting any product subset with positive Lebesgue mea-sure. Note also that W∞ is a “square-like” region with area equal to a circle

17

with radius 12. In that sense, this is a solution to the ancient problem of

“squaring the circle”.

Sources

This section is adapted from a sketch of the proof of Wallis’ Formula inKazarinoff [3] and the short note by Wastlund [8]. The elementary proofs ofthe sine product are from [4] and [2]. The geometric interpretation of Wallis’Formula is adapted from [6].

Problems to Work for Understanding

1. Show that

24n · (n!)4

((2n)!)2(2n+ 1)=

2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n+ 1).

2. Numerically find the rate at which the sequence

2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)

1 · 3 · 3 · 5 · 5 . . . (2n− 1) · (2n+ 1)

approaches π/2 by calculating values and using regression.

3. Use induction to prove

J2n+1 =(2n) · (2n− 2) . . . 4 · 2

(2n+ 1) · (2n− 1) . . . 3 · 1

and

J2n =(2n− 1) · (2n− 3) . . . 3 · π(2n) · (2n− 2) . . . 4 · 2 · 2 .

18

4. The simple proof of the sine product formula uses the polynomial Fmof degree m such that

sin((2m+ 1)x) = sin(x) · Fm(sin2(x)).

Explicitly find F0(x), F1(x), F2(x), F3(x), F4(x) and plot them on [−1, 1].

5. Give a detailed proof that

∫ n

0

xt−1(

1− x

n

)ndx =

n!nt

t(t+ 1) . . . (t+ n).

6. Provide a careful and detailed proof that if X and Y are independentexponential random variables, both with expectation 1 we have that

E[(X/Y )t

]= E

[(X/Y )|t|

].

7. Provide a detailed proof using contour integration in the complex planeto show that

∫ ∞

0

|t|u|t|−1u+ 1

du =|t|2πieπi |t|

e2πi|t| − 1=

πt

sin(πt).

Reading Suggestion:

References

[1] Michael D. Hirschhorn. Wallis’s product and the central binomial coef-ficient. American Mathematical Monthly, 122(7):689, August-September2015.

[2] Lars Holst. A proof of Euler’s infinite product for the sine. AmericanMathematical Monthly, 119:518–521, June-July 2012.

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[3] Nicholaus D. Kazarinoff. Analytic Inequalities. Holt, Rinehart and Win-ston, 1961.

[4] R. A. Kortram. Simple proofs for∑∞

k=11k2

= π2

6and sinx =

x∏∞

k=1

(1− x2

k2π2

). Mathematics Magazine, 69(2):123–125, April 1996.

[5] Oscar Ciaurri. Euler’s product expansion for the sine: An elemen-tary proof. American Mathematical Monthly, 122(7):693–695, August-September 2015.

[6] Hansklaus Rummler. Squaring the circle with holes. American Mathe-matical Monthly, 100(9):858–860, November 1993.

[7] S. Saks and A. Zygmund. Analytic Functions. Elsevier Publishing, 1971.

[8] Johan Wastlund. An elementary proof of the wallis product formula forpi. American Mathematical Monthly, 114(10):914–917, December 2007.

Outside Readings and Links:

1.

2.

3.

4.

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