Steven R. Dunbar
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar

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Wallis’ Formula

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Rating

Rating

Mathematically Mature: may contain mathematics beyond calculus with proofs.

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Section Starter Question

Section Starter Question

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

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Key Concepts

Key Concepts

  1. Wallis’ Formula is the amazing limit
    lim n 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 from integration of trigonometric functions.
  3. Another proof uses only basic algebra, the Pythagorean Theorem, and the formula πr2 for the area of a circle of radius r.
  4. Yet another proof uses Euler’s infinite product representation for the sine function.

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Vocabulary

Vocabulary

  1. Wallis’ Formula is the amazing limit
    lim n 2 2 4 4 6 6(2n) (2n) 1 3 3 5 5(2n 1) (2n 1) (2n + 1) = π 2.

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Mathematical Ideas

Mathematical Ideas

Introduction

Wallis’ Formula is the amazing limit

lim n 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

lim n 24n (n!)4 ((2n)!)2(2n + 1) = π 2

or equivalently

lim n 24n 2n n 2(2n + 1) = π 2.

Note that Wallis Formula is equivalent to saying that the “central binomial 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 inequality.

In the form

wn = j=1n (2j)2 (2j 1)(2j + 1) = 24n 2n n 2(2n + 1) (2)

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


Convergence of the Wallis formula

Figure 1: Convergence of the Wallis formula to π2.

Doing a numerical linear regression of log(π2 wn) versus log n on the domain n = 1 to n = 30 indicates that wn approaches π2 at a rate which is O(1n).

A proof using integration and recursion

Theorem 1 (Wallis’ Formula).

lim n 2 2 4 4 6 6(2n) (2n) 1 3 3 5 5(2n 1) (2n + 1) = π 2. (3)

Proof. Consider Jn =0π2 cos n(x)dx. Integrating by parts with u = cos n1(x) and dv = cos(x) shows

0π2 cos n(x)dx = (n 1)0π2 cos n2(x) sin 2(x)dx = (n 1)0π2 cos n2(x)(1 cos 2(x))dx = (n 1)0π2 cos n2(x)dx (n 1)0π2 cos n(x)dx.

Gathering terms, we get nJn = (n 1)Jn2.

Now J1 = 1 so recursively J3 = 2 3, J5 = 24 35 and inductively

J2n+1 = 2 4(2n 2) (2n) 1 3(2n 1) (2n + 1). (4)

Likewise J2 = π 22, and J4 = 3π 242, J6 = 35π 2462 and inductively

J2n = 3 5(2n 3) (2n 1) π 2 4(2n 2) (2n) 2 .

For 0 x π2, 0 cos(x) 1, so cos 2n(x) cos 2n+1(x) cos 2n+2(x), implying in turn that J2n J2n+1 J2n+2. Then

1 J2n+1 J2n J2n+2 J2n = 2n + 1 2n + 2.

Hence lim nJ2n+1 J2n = 1.

That is,

lim n 2 2 4 4 6 6(2n) (2n) 1 3 3 5 5(2n 1) (2n + 1) 2 π = 1

or equivalently

lim n 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 only basic algebra, the Pythagorean Theorem, and the formula πr2 for the area of a circle of radius r. Another important property used implicitly is the completeness property of the reals.

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

sn = 3 2 5 42n 1 2n 2. (5)

These are the reciprocals of the subsequence J2n1 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 sn2, (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 sn2 . (7)

Interpret e1 = 1 as an empty product. Since (2n)2 (2n1)(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

for any n. Furthermore, for 1 i n,

2i si2 = oi on

and

2i 1 si2 = ei en

from which it follows that

2i 1 en si2 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 for n 1

an = sn+1 sn = sn 2n + 1 2n 1 = sn 2n = 1 2 3 42n 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.

Lemma 3.

1 = a02 = a 0a1 + a1a0 = a0a2 + a12 + a 2a0 = a0an + a1an1 + + ana0

Proof. Start from a02 = 1 and repeatedly apply the identity (9). At stage n applying the identity (9) to every term the sum

a0an + a1an1 + + ana0

becomes

a0an + 1 na1an1 + n 1 n a1an1 + 2 na2an2 + + 1 nan1a1 + ana0 .

Collecting terms, this simplifies to a0an + a1an1 + + ana0. □

Now divide the positive quadrant of the xy-plane into rectangles by drawing the vertical lines y = sn and the horizontal lines y = sn for all n. Let Ri,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 = n is 1. Let Pn be the polygonal region consisting of all rectangles Ri,j for which i + 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 and 1 i,j, n. By the Pythagorean theorem, the distance of such a point to the origin is si2 + sj2. By (8)

2(i + j) on = 2(n + 1) on .

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

2(i + j 1) en = 2(n 1) en .

Therefore, Pn contains a quarter circle of radius 2(n 1)en and is contained in a quarter circle of radius 2(n + 1)on. See Figure 2 for a diagram of the polygonal region Pn and the corresponding inner and outer quarter circles for n = 4.


Diagram of the regions

Figure 2: A diagram of the regions Ri,j and the inner and outer quarter circles for the case 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 π2.

Wallis’ Formula and the Central Binomial Coefficient

This subsection gives a detailed proof that Wallis’ Formula gives an explicit inequality bound on the central binomial term that in turn implies the asymptotic formula for the central binomial coefficient. The derivation in [1]. motivates this proof.

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

wn = j=1n (2j)2 (2j 1)(2j + 1) = 24n 2n n 2(2n + 1).

Rearrange it to

2n n 2 = 16n 2n + 1 j=1n(2j 1)(2j + 1) (2j)2

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

2n n 2 = 16n (2n + 1) 2n + 2 (2n + 1)on+1

and

2n n 2 = 16n (2n + 1) 1 en+1.

Rearrange the inequality (10) to obtain

2n + 2 π(n + 2) < 1 on+1 and  1 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 expressions 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=11 z2 m2π2 .

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

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

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

cos(2kx) = Tk(cos(2x)) = Tk(1 2 sin 2x).

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(sin 2(x)).

(In fact, using the definition Un(cos(x)) = sin((n+1)x) sin(x) for the Chebyshev polynomials of the second kind, it is easy to show that Fm(x) = U2n(1 x).) Substituting xk = kπ 2m+1 and noting that sin((2m + 1)xk) = 0 shows that Fm has zeros at sin 2 ( kπ 2m+1 for k = 1, 2,,m. These zeros are distinct, so Fm has no other zeros, then

Fm(y) = Fm(0) k=1m 1 y sin 2 kπ 2m+1

and

Fm(0) = lim x0 sin((2m + 1)x) sin(x) = 2m + 1.

Therefore

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

and changing variables

sin(x) = (2m + 1) sin x 2m + 1 k=1m 1 sin 2 x 2m+1 sin 2 kπ 2m+1 . (11)

The goal now is to estimate the product terms. For all real t, et 1 + t and therefore et 1 1+t. For u < 1, the choice t = u(1 u) gives eu(1u) 1 u. Then for every collection of numbers uk [0, 1), we have

1 k uk 1 uk e k uk 1uk k(1 uk) e kuk 1. (12)

If also kuk < 1, then we also know that

e kuk 1 1 + kuk (13)

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

kuk 1 + kuk 1 k(1 uk) k uk 1 uk. (14)

Let m and N be positive integers with m > N. Take x such that |x| < 1 4Nπ and x π. Then define uk by

uk = sin x 2m+1 sin kπ 2m+1 2,k = 1, 2,,m.

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

sin(x) (2m + 1) sin x 2m+1 k=1N(1 uk) = k=N+1m(1 u k).

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

1 k=N+1m uk 1 uk sin(x) (2m + 1) sin x 2m+1 k=1N(1 uk) 1. (15)

For 0 t π 2 we have sin(t) 2 πt, so we see that

uk (2m + 1) sin x 2m+1 2k 2 x 2k2;

thus

uk 1 uk x2 (2k)2 x2fork > N.

Hence

k=N+1m uk 1 uk x2 2N |x|.

Thus it follows from (15) that

1 x2 2N |x| sin(x) (2m + 1) sin x 2m+1 k=1N(1 uk) 1.

Let m so that

1 x2 2N |x| sin(x) x k=1N(1 uk) 1.

Now let N and we obtain for x π

sin x = x k=11 x2 k2π2 .

For x π this equality is also true. □

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

Proof. Start with the integral

0nxt1 1 x nndx

and integrate by parts once:

u = 1 x nn v = 1 txt du = 1 x nn1dx dv = xt1dx.

Then

0nxt1 1 x nndx = 1 x nn 0n +0n1 txt 1 x nn1dx = 1 t0nxt 1 x nn1dx

Integrate by parts again:

u = 1 x nn1 v = 1 t + 1xt+1 du = n 1 n 1 x nn2dx dv = xtdx

so

0nxt1 1 x nndx = (n 1) t(t + 1)n0nxt+1 1 x nn2dx.

After integration by parts n times:

0nxt1 1 x nndx = (n 1)! t(t + 1)(t + n 1)nn10nxt+n1dx = (n 1)! t(t + 1)(t + n 1)nn1 xt+n t + n0n = n!nt t(t + 1)(t + n).

Then by dominated convergence it follows that

Γ(t) =0xt1exdx = lim n n!nt t(t + 1)(t + n).

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

Γ(1 + t)Γ(1 t) = k=1 1 1 t2k2.

For X and Y independent exponential random variables, both with expectation 1 we have that

𝔼 Xt =0xtexdx = Γ(1 + t)

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

Γ(1 + t)Γ(1 t) = 𝔼 Xt 𝔼 Y t = 𝔼 (XY )t = 𝔼 (XY )|t|

The distribution function for the random variable XY is

XY u = Y Xu = 𝔼 eXu = u u + 1

for u > 0. Then the probability density is

d du XY u = 1 (1 + u)2,u > 0.

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

𝔼 (XY )|t| =0 u|t| (u + 1)2du =0|t|u|t|1 u + 1 du

The last integral is πt sin(πt). For example, this is formula 613, page 445 in the 18th Edition of the CRC Standard Mathematical Tables. It can also be done with contour integration in the complex plane

0+|t|u|t|1 u + 1 du 2πi(1)|t|1 ++0|t|u2πi(|t|1) u + 1 du = 0

so

0|t|u|t|1 u + 1 du 1 e2πi|t| = 2πieπi|t|.

Thus

0|t|u|t|1 u + 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 (XY )t = πt sin(πt).

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

sin(πt) = π Γ(t)Γ(1 t).

Combining all results above,

sin(πt) = πt k=11 t2 k2 .

Corollary 1 (Wallis’ Formula).

lim n 2 2 4 4 6 6(2n) (2n) 1 3 3 5 5(2n 1) (2n 1) (2n + 1) = π 2.

Proof. Substitute z = π2 in the product expansion of sin(z). □

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

A Geometric Interpretation of Wallis’ Formula

This section gives a geometric interpretation of Wallis’ formula, constructing a subset of the unit square with area 8 9 24 25 48 49 = 2 3 4 3 4 5 6 5 6 7 8 7 = 1 2 π 2 . The construction is reminiscent of the construction of the Sierpinski Triangle.

Take a unit square W0 and remove a square of area 1 3 1 3 from the center. Label the result as W1 with area 8 9. Next remove a square of area 1 5 1 5 from the center of each of the 8 small sub-squares of area 1 9 constituting W1 and call the result W2 with area 8 9 24 25. Continuing by removing squares of area 1 7 1 7 from each sub-square of area 1 5 1 5 surrounding the hole removed at stage 2, construct region W3 with area 8 9 24 25 48 49. Continue to create W4, and so on. Call the limiting region W which has area 8 9 24 25 48 49 = 1 2 π 2 .

The side length of the removed squares at stage n is 2nn! (2n+1)!. After 3 stages, 201 squares have been removed, and the stage 3 holes have side length 1 3 1 5 1 7 = 1 105. It will be impractical to illustrate more than 3 stages of this iterative process.


Three stages in the construction of the
			       Wallis sieve

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

Here area means the Lebesgue measure of the limiting region W. However W does not contain any product set A × B with A,B , each with positive Lebesgue measure. To see this, consider a maximal product subset of 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 2 3 4 5. Continuing, a maximal product subset of W has measure 2 3 4 5 6 7 = n=11 1 2n+1 = 0.

Thus, W is an example of a subset of the plane 2 with positive Lebesgue measure, but not admitting any product subset with positive Lebesgue measure. Note also that W is a “square-like” region with area equal to a circle with radius 1 2. 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 in Kazarinoff [3] and the short note by Wästlund [8]. The elementary proofs of the sine product are from [4] and [2]. The geometric interpretation of Wallis’ Formula is adapted from [6].

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Problems to Work

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 π2 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 .

  4. The simple proof of the sine product formula uses the polynomial Fm of degree m such that
    sin((2m + 1)x) = sin(x) Fm(sin 2(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
    0nxt1 1 x nndx = n!nt t(t + 1)(t + n).

  6. Provide a careful and detailed proof that if X and Y are independent exponential random variables, both with expectation 1 we have that
    𝔼 (XY )t = 𝔼 (XY )|t|.

  7. Provide a detailed proof using contour integration in the complex plane to show that
    0|t|u|t|1 u + 1 du = |t|2πieπi|t| e2πi|t| 1 = πt sin(πt).

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Books

Reading Suggestion:

References

[1]   Michael D. Hirschhorn. Wallis’s product and the central binomial coefficient. American Mathematical Monthly, 122(7):689, August-September 2015.

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

[3]   Nicholaus D. Kazarinoff. Analytic Inequalities. Holt, Rinehart and Winston, 1961.

[4]   R. A. Kortram. Simple proofs for k=1 1 k2 = π2 6 and sin x = x k=11 x2 k2π2. Mathematics Magazine, 69(2):123–125, April 1996.

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

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

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

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

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Links

Outside Readings and Links:

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