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.

19

[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.

I check all the information on each page for correctness and typographicalerrors. Nevertheless, some errors may occur and I would be grateful if you wouldalert me to such errors. I make every reasonable effort to present current andaccurate information for public use, however I do not guarantee the accuracy ortimeliness of information on this website. Your use of the information from thiswebsite is strictly voluntary and at your risk.

20

I have checked the links to external sites for usefulness. Links to externalwebsites are provided as a convenience. I do not endorse, control, monitor, orguarantee the information contained in any external website. I don’t guaranteethat the links are active at all times. Use the links here with the same caution asyou would all information on the Internet. This website reflects the thoughts, in-terests and opinions of its author. They do not explicitly represent official positionsor policies of my employer.

Information on this website is subject to change without notice.

Steve Dunbar’s Home Page, http://www.math.unl.edu/~sdunbar1Email to Steve Dunbar, sdunbar1 at unl dot edu

Last modified: Processed from LATEX source on November 2, 2017

21