topics in probability theory and stochastic processes steven r
TRANSCRIPT
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:
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4.
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