Wallis\' Formula and Other Infinite Products

Wallis’ Formula and Other Infinite Products Richard Belshoff Missouri State University MAA Student Chapter Talk

January 29, 2009

Wallis’ formula

π 2 2 4 4 6 6 = · · · · · · ··· 2 1 3 3 5 5 7

Wallis’ formula

π 2 2 4 4 6 6 = · · · · · · ··· 2 1 3 3 5 5 7     2 2 4 4 6 6 π · · · ··· = . 1 3 3 5 5 7 2

Wallis’ formula

π 2 2 4 4 6 6 = · · · · · · ··· 2 1 3 3 5 5 7     2 2 4 4 6 6 π · · · ··· = . 1 3 3 5 5 7 2 Now pull out every other pair of terms.

Wallis’ formula

π 2 2 4 4 6 6 = · · · · · · ··· 2 1 3 3 5 5 7     2 2 4 4 6 6 π · · · ··· = . 1 3 3 5 5 7 2 Now pull out every other pair of terms.     2 2 6 6 10 10 · · · ··· = 1 3 5 7 9 11

Wallis’ formula

π 2 2 4 4 6 6 = · · · · · · ··· 2 1 3 3 5 5 7     2 2 4 4 6 6 π · · · ··· = . 1 3 3 5 5 7 2 Now pull out every other pair of terms.     √ 2 2 6 6 10 10 · · · ··· = 2 1 3 5 7 9 11

Wallis’ formula

π 2 2 4 4 6 6 = · · · · · · ··· 2 1 3 3 5 5 7     2 2 4 4 6 6 π · · · ··· = . 1 3 3 5 5 7 2 Now pull out every other pair of terms.     √ 2 2 6 6 10 10 · · · ··· = 2 1 3 5 7 9 11

AMAZING!

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products”

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 ,

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 ,

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 ,

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . ,

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an ,

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an , . . . . . . .

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an , . . . . . . . If pn → M as n → ∞, i.e. if lim pn = M, n→∞

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an , . . . . . . . If pn → M as n → ∞, i.e. if lim pn = M, then we say that n→∞

a1 a2 a3 · · · =M.

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an , . . . . . . . If pn → M as n → ∞, i.e. if lim pn = M, then we say that n→∞

a1 a2 a3 · · · =M. Notation:

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an , . . . . . . . If pn → M as n → ∞, i.e. if lim pn = M, then we say that n→∞

a1 a2 a3 · · · =M. Notation:

∞ Y i=1

ai = a1 a2 a3 · · ·

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an , . . . . . . . If pn → M as n → ∞, i.e. if lim pn = M, then we say that n→∞

a1 a2 a3 · · · =M. Notation:

∞ Y

ai = a1 a2 a3 · · ·

i=1 n Y i=1

ai = a1 a2 · · · an

What do we mean by an infinite product? ?

a1 a2 a3 · · · = M Consider the sequence of “partial products” p1 = a1 , p2 = a1 a2 , p3 = a1 a2 a3 , . . . , pn = a1 · · · an , . . . . . . . If pn → M as n → ∞, i.e. if lim pn = M, then we say that n→∞

a1 a2 a3 · · · =M. Notation:

∞ Y

ai = a1 a2 a3 · · ·

i=1 n Y

ai = a1 a2 · · · an

i=1 ∞ Y i=1

ai = lim

n→∞

n Y i=1

ai

The Factor Theorem Theorem (Factor Theorem) A real number r is a root of a polynomial p(x) if and only if (x − r ) is a factor of p(x).

The Factor Theorem Theorem (Factor Theorem) A real number r is a root of a polynomial p(x) if and only if (x − r ) is a factor of p(x).

Example A polynomial with roots −2, −1, 0, 1, and 2 is

The Factor Theorem Theorem (Factor Theorem) A real number r is a root of a polynomial p(x) if and only if (x − r ) is a factor of p(x).

Example A polynomial with roots −2, −1, 0, 1, and 2 is p(x) = (x + 2)(x + 1)x(x − 1)(x − 2)

The Factor Theorem Theorem (Factor Theorem) A real number r is a root of a polynomial p(x) if and only if (x − r ) is a factor of p(x).

Example A polynomial with roots −2, −1, 0, 1, and 2 is p(x) = (x + 2)(x + 1)x(x − 1)(x − 2) = x(x 2 − 1)(x 2 − 4).

The Factor Theorem Theorem (Factor Theorem) A real number r is a root of a polynomial p(x) if and only if (x − r ) is a factor of p(x).

Example A polynomial with roots −2, −1, 0, 1, and 2 is p(x) = (x + 2)(x + 1)x(x − 1)(x − 2) = x(x 2 − 1)(x 2 − 4). But this is not unique

The Factor Theorem Theorem (Factor Theorem) A real number r is a root of a polynomial p(x) if and only if (x − r ) is a factor of p(x).

Example A polynomial with roots −2, −1, 0, 1, and 2 is p(x) = (x + 2)(x + 1)x(x − 1)(x − 2) = x(x 2 − 1)(x 2 − 4). But this is not unique p(x) = Cx(x 2 − 1)(x 2 − 4)

An infinite number of roots? What about an infinite number of roots?

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .?

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .? ? f (x) = Cx(x 2 − 1)(x 2 − 4)(x 2 − 9) · · ·

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .? ? f (x) = Cx(x 2 − 1)(x 2 − 4)(x 2 − 9) · · · Better:

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .? ? f (x) = Cx(x 2 − 1)(x 2 − 4)(x 2 − 9) · · · Better: f (x) = Cx(1 − x 2 )(1 −

x2 x2 )(1 − ) · · · 4 9

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .? ? f (x) = Cx(x 2 − 1)(x 2 − 4)(x 2 − 9) · · · Better:

x2 x2 )(1 − ) · · · 4 9 This infinite product converges for all values of x f (x) = Cx(1 − x 2 )(1 −

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .? ? f (x) = Cx(x 2 − 1)(x 2 − 4)(x 2 − 9) · · · Better:

x2 x2 )(1 − ) · · · 4 9 This infinite product converges for all values of x (Knopp, Theory of Functions, Part II, Dover, New York, 1947.) f (x) = Cx(1 − x 2 )(1 −

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .? ? f (x) = Cx(x 2 − 1)(x 2 − 4)(x 2 − 9) · · · Better:

x2 x2 )(1 − ) · · · 4 9 This infinite product converges for all values of x (Knopp, Theory of Functions, Part II, Dover, New York, 1947.) f (x) = Cx(1 − x 2 )(1 −

There is a well-known function that behaves similarly, that is zero at all the integers, namely, .....

An infinite number of roots? What about an infinite number of roots? Could we construct an “infinite polynomial” with roots 0, ±1, ±2, ±3, . . .? ? f (x) = Cx(x 2 − 1)(x 2 − 4)(x 2 − 9) · · · Better:

x2 x2 )(1 − ) · · · 4 9 This infinite product converges for all values of x (Knopp, Theory of Functions, Part II, Dover, New York, 1947.) f (x) = Cx(1 − x 2 )(1 −

There is a well-known function that behaves similarly, that is zero at all the integers, namely, ..... sin πx.

I

f (x) = Cx(1 − x 2 )(1 −

x2 4 )(1

−

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

x2 4 )(1

−

x2 9 )···

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

−

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

−

x2 9 )···

= sin πx, then C = π.

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

C (1 − x 2 )(1 −

−

x2 9 )···

= sin πx, then C = π.

x2 x2 )(1 − ) · · · = 4 9

sin πx x

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

lim C (1 − x 2 )(1 −

x→0

−

x2 9 )···

= sin πx, then C = π.

x2 sin πx x2 )(1 − ) · · · = lim x→0 4 9 x

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

lim C (1 − x 2 )(1 −

x→0

Therefore C = π.

−

x2 9 )···

= sin πx, then C = π.

x2 sin πx x2 )(1 − ) · · · = lim x→0 4 9 x

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

lim C (1 − x 2 )(1 −

x→0

−

x2 9 )···

= sin πx, then C = π.

x2 sin πx x2 )(1 − ) · · · = lim x→0 4 9 x

Therefore C = π. I

When C = π, we have f (x) = sin πx for all values of x.

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

lim C (1 − x 2 )(1 −

x→0

−

x2 9 )···

= sin πx, then C = π.

x2 sin πx x2 )(1 − ) · · · = lim x→0 4 9 x

Therefore C = π. I

When C = π, we have f (x) = sin πx for all values of x.

I

Euler’s infinite product for sine:

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

lim C (1 − x 2 )(1 −

x→0

−

x2 9 )···

= sin πx, then C = π.

x2 sin πx x2 )(1 − ) · · · = lim x→0 4 9 x

Therefore C = π. I

When C = π, we have f (x) = sin πx for all values of x.

I

Euler’s infinite product for sine: sin πx = πx(1 − x 2 )(1 −

x2 x2 )(1 − ) · · · 4 9

x2 4 )(1

x2 9 )···

I

f (x) = Cx(1 − x 2 )(1 −

I

sin πx

I

Both are zero at all the integers.

I

If Cx(1 − x 2 )(1 −

x2 4 )(1

−

lim C (1 − x 2 )(1 −

x→0

−

x2 9 )···

= sin πx, then C = π.

x2 sin πx x2 )(1 − ) · · · = lim x→0 4 9 x

Therefore C = π. I

When C = π, we have f (x) = sin πx for all values of x.

I

Euler’s infinite product for sine: sin πx = πx(1 − x 2 )(1 −

AMAZING!

x2 x2 )(1 − ) · · · 4 9

Digression: The Basel Problem

Digression: The Basel Problem

Aside: Euler used this identity

Digression: The Basel Problem

Aside: Euler used this identity sin πx = πx(1 − x 2 )(1 −

x2 x2 )(1 − ) · · · 4 9

Digression: The Basel Problem

Aside: Euler used this identity sin πx = πx(1 − x 2 )(1 − to prove

x2 x2 )(1 − ) · · · 4 9

Digression: The Basel Problem

Aside: Euler used this identity sin πx = πx(1 − x 2 )(1 −

x2 x2 )(1 − ) · · · 4 9

to prove 1+

1 1 1 π2 . + 2 + 2 + ··· = 2 2 3 4 6

Digression: The Basel Problem

Aside: Euler used this identity sin πx = πx(1 − x 2 )(1 −

x2 x2 )(1 − ) · · · 4 9

to prove 1+

1 1 1 π2 . + 2 + 2 + ··· = 2 2 3 4 6

Our goal: prove Wallis’ formula.

Proof of Wallis’ Formula

Proof of Wallis’ Formula

sin πx

= πx(1 − x 2 )(1 −

x2 x2 )(1 − ) · · · 4 9

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 π 3 15 35 sin(π/2) = · ··· 2 4 16 36 sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 π 1 = 2 sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 π 1·3 1 = · 2 2·2 sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 π 1·3 3·5 1 = · · 2 2·2 4·4 sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 π 1·3 3·5 5·7 1 = · · · 2 2·2 4·4 6·6 sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 π 1·3 3·5 5·7 1 = · · · ········· 2 2·2 4·4 6·6 sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 π 1·3 3·5 5·7 1 = · · · ········· 2 2·2 4·4 6·6 Therefore, sin πx

Proof of Wallis’ Formula

x2 x2 = πx(1 − x 2 )(1 − )(1 − ) · · ·     4  9 (2n − 1)(2n + 1) π 3 15 35 sin(π/2) = · ··· ··· 2 4 16 36 (2n)2 π 1·3 3·5 5·7 1 = · · · ········· 2 2·2 4·4 6·6 Therefore, π 2 2 4 4 6 6 = · · · · · · ··· 2 1 3 3 5 5 7 sin πx

An infinite product for cosine

An infinite product for cosine sin(πx) = πx(1 − x 2 )(1 −

x2 x2 x2 x2 )(1 − )(1 − )(1 − ) · · · 4 9 16 25

An infinite product for cosine x2 x2 x2 x2 )(1 − )(1 − )(1 − ) · · · 4 9 16 25 2 2 2 4x 4x 4x 4x 2 sin(2πx) = 2πx(1 − 4x 2 )(1 − )(1 − )(1 − )(1 − )··· 4 9 16 25 sin(πx) = πx(1 − x 2 )(1 −

An infinite product for cosine x2 x2 x2 x2 )(1 − )(1 − )(1 − ) · · · 4 9 16 25 2 2 2 4x 4x 4x 4x 2 sin(2πx) = 2πx(1 − 4x 2 )(1 − )(1 − )(1 − )(1 − )··· 4 9 16 25 4x 2 x2 4x 2 sin(2πx) = 2πx(1 − 4x 2 )(1 − x 2 )(1 − )(1 − )(1 − )··· 9 4 25 sin(πx) = πx(1 − x 2 )(1 −

An infinite product for cosine

x2 x2 x2 x2 )(1 − )(1 − )(1 − ) · · · 4 9 16 25 2 2 2 4x 4x 4x 4x 2 sin(2πx) = 2πx(1 − 4x 2 )(1 − )(1 − )(1 − )(1 − )··· 4 9 16 25 4x 2 x2 4x 2 sin(2πx) = 2πx(1 − 4x 2 )(1 − x 2 )(1 − )(1 − )(1 − )··· 9 4 25    4x 2 4x 2 x2 )(1 − ) sin(2πx) = 2 πx(1 − x 2 )(1 − ) · · · (1 − 4x 2 )(1 − 4 9 25 sin(πx) = πx(1 − x 2 )(1 −

An infinite product for cosine

x2 x2 x2 x2 )(1 − )(1 − )(1 − ) · · · 4 9 16 25 2 2 2 4x 4x 4x 4x 2 2πx(1 − 4x 2 )(1 − )(1 − )(1 − )(1 − )··· 4 9 16 25 4x 2 x2 4x 2 2πx(1 − 4x 2 )(1 − x 2 )(1 − )(1 − )(1 − )··· 9 4 25    4x 2 4x 2 x2 )(1 − ) 2 πx(1 − x 2 )(1 − ) · · · (1 − 4x 2 )(1 − 4 9 25 2 sin(πx) cos(πx)

sin(πx) = πx(1 − x 2 )(1 − sin(2πx) = sin(2πx) = sin(2πx) = sin(2πx) =

An infinite product for cosine

x2 x2 x2 x2 )(1 − )(1 − )(1 − ) · · · 4 9 16 25 2 2 2 4x 4x 4x 4x 2 2πx(1 − 4x 2 )(1 − )(1 − )(1 − )(1 − )··· 4 9 16 25 4x 2 x2 4x 2 2πx(1 − 4x 2 )(1 − x 2 )(1 − )(1 − )(1 − )··· 9 4 25    4x 2 4x 2 x2 )(1 − ) 2 πx(1 − x 2 )(1 − ) · · · (1 − 4x 2 )(1 − 4 9 25 2 sin(πx) cos(πx)

sin(πx) = πx(1 − x 2 )(1 − sin(2πx) = sin(2πx) = sin(2πx) = sin(2πx) =

Therefore,

An infinite product for cosine

x2 x2 x2 x2 )(1 − )(1 − )(1 − ) · · · 4 9 16 25 2 2 2 4x 4x 4x 4x 2 2πx(1 − 4x 2 )(1 − )(1 − )(1 − )(1 − )··· 4 9 16 25 4x 2 x2 4x 2 2πx(1 − 4x 2 )(1 − x 2 )(1 − )(1 − )(1 − )··· 9 4 25    4x 2 4x 2 x2 )(1 − ) 2 πx(1 − x 2 )(1 − ) · · · (1 − 4x 2 )(1 − 4 9 25 2 sin(πx) cos(πx)

sin(πx) = πx(1 − x 2 )(1 − sin(2πx) = sin(2πx) = sin(2πx) = sin(2πx) =

Therefore, cos(πx) = (1 − 4x 2 )(1 −

4x 2 4x 2 )(1 − )··· 9 25

The infinite product for

√

2

The infinite product for

√

2

cos(πx) = (1 − 4x 2 )(1 −

4x 2 4x 2 )(1 − )··· 9 25

The infinite product for

√

2

4x 2 4x 2 )(1 − )··· cos(πx) = (1 − 4x 2 )(1 − 25    9  3 35 99 cos(π/4) = ··· 4 36 100

The infinite product for

√

2

4x 2 4x 2 )(1 − )··· cos(πx) = (1 − 4x 2 )(1 − 25    9  3 35 99 cos(π/4) = ··· 4 36 100     5·7 9 · 11 1 1·3 √ ··· = 2·2 6·6 10 · 10 2

The infinite product for

√

2

4x 2 4x 2 )(1 − )··· cos(πx) = (1 − 4x 2 )(1 − 25    9  3 35 99 cos(π/4) = ··· 4 36 100     5·7 9 · 11 1 1·3 √ ··· = 2·2 6·6 10 · 10 2 Therefore,

The infinite product for

√

2

4x 2 4x 2 )(1 − )··· cos(πx) = (1 − 4x 2 )(1 − 25    9  3 35 99 cos(π/4) = ··· 4 36 100     5·7 9 · 11 1 1·3 √ ··· = 2·2 6·6 10 · 10 2 Therefore,     √ 2 2 6 6 10 10 · · · 2 = ··· 1 3 5 7 9 11

That’s all folks!

THE END

References

Su, Francis E., et al. ”Wallis’ Formula.” Mudd Math Fun Facts. http://www.math.hmc.edu/funfacts.

References

Su, Francis E., et al. ”Wallis’ Formula.” Mudd Math Fun Facts. http://www.math.hmc.edu/funfacts. Vandervelde, S., “Newton’s Sums and the Infinite Product Representation for sin πx,” Mathematics and Informatics Quarterly, 9 (1999), pp. 64-69.