richard belsho missouri state university maa student
TRANSCRIPT
Wallis’ Formula and Other Infinite Products
Richard BelshoffMissouri State University
MAA Student Chapter Talk
January 29, 2009
Wallis’ formula
π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
(2
1· 2
3
)(4
3· 4
5
)(6
5· 6
7
)· · · =
π
2.
Now pull out every other pair of terms.(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · · =
√2
AMAZING!
Wallis’ formula
π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
(2
1· 2
3
)(4
3· 4
5
)(6
5· 6
7
)· · · =
π
2.
Now pull out every other pair of terms.(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · · =
√2
AMAZING!
Wallis’ formula
π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
(2
1· 2
3
)(4
3· 4
5
)(6
5· 6
7
)· · · =
π
2.
Now pull out every other pair of terms.
(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · · =
√2
AMAZING!
Wallis’ formula
π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
(2
1· 2
3
)(4
3· 4
5
)(6
5· 6
7
)· · · =
π
2.
Now pull out every other pair of terms.(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · · =
√2
AMAZING!
Wallis’ formula
π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
(2
1· 2
3
)(4
3· 4
5
)(6
5· 6
7
)· · · =
π
2.
Now pull out every other pair of terms.(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · · =
√2
AMAZING!
Wallis’ formula
π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
(2
1· 2
3
)(4
3· 4
5
)(6
5· 6
7
)· · · =
π
2.
Now pull out every other pair of terms.(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · · =
√2
AMAZING!
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”
p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1,
p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2,
p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3,
. . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . ,
pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an,
. . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M,
then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:
∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏i=1
ai
What do we mean by an infinite product?
a1a2a3 · · ·?= M
Consider the sequence of “partial products”p1 = a1, p2 = a1a2, p3 = a1a2a3, . . . , pn = a1 · · · an, . . . . . . .
If pn → M as n→∞, i.e. if limn→∞
pn = M, then we say that
a1a2a3 · · ·=M.
Notation:∞∏i=1
ai = a1a2a3 · · ·
n∏i=1
ai = a1a2 · · · an
∞∏i=1
ai = limn→∞
n∏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).
Example
A polynomial with roots −2, −1, 0, 1, and 2 is
p(x) = (x + 2)(x + 1)x(x − 1)(x − 2)
= x(x2 − 1)(x2 − 4).
But this is not unique
p(x) = Cx(x2 − 1)(x2 − 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(x2 − 1)(x2 − 4).
But this is not unique
p(x) = Cx(x2 − 1)(x2 − 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(x2 − 1)(x2 − 4).
But this is not unique
p(x) = Cx(x2 − 1)(x2 − 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(x2 − 1)(x2 − 4).
But this is not unique
p(x) = Cx(x2 − 1)(x2 − 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(x2 − 1)(x2 − 4).
But this is not unique
p(x) = Cx(x2 − 1)(x2 − 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(x2 − 1)(x2 − 4).
But this is not unique
p(x) = Cx(x2 − 1)(x2 − 4)
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x
(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, .....
sinπx .
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(x2 − 1)(x2 − 4)(x2 − 9) · · ·
Better:
f (x) = Cx(1− x2)(1− x2
4)(1− x2
9) · · ·
This infinite product converges for all values of x(Knopp, Theory of Functions, Part II, Dover, New York, 1947.)
There is a well-known function that behaves similarly, that is zeroat all the integers, namely, ..... sinπx .
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · =
limx→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·
I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · =
limx→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · =
limx→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · =
limx→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · =
limx→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · =
limx→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · = lim
x→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · = lim
x→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · = lim
x→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · = lim
x→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · = lim
x→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
I f (x) = Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · ·I sinπx
I Both are zero at all the integers.
I If Cx(1− x2)(1− x2
4 )(1− x2
9 ) · · · = sinπx , then C = π.
limx→0
C (1− x2)(1− x2
4)(1− x2
9) · · · = lim
x→0
sinπx
x
Therefore C = π.
I When C = π, we have f (x) = sinπx for all values of x .
I Euler’s infinite product for sine:
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
AMAZING!
Digression: The Basel Problem
Aside: Euler used this identity
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
to prove
1 +1
22+
1
32+
1
42+ · · · =
π2
6.
Our goal: prove Wallis’ formula.
Digression: The Basel Problem
Aside: Euler used this identity
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
to prove
1 +1
22+
1
32+
1
42+ · · · =
π2
6.
Our goal: prove Wallis’ formula.
Digression: The Basel Problem
Aside: Euler used this identity
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
to prove
1 +1
22+
1
32+
1
42+ · · · =
π2
6.
Our goal: prove Wallis’ formula.
Digression: The Basel Problem
Aside: Euler used this identity
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
to prove
1 +1
22+
1
32+
1
42+ · · · =
π2
6.
Our goal: prove Wallis’ formula.
Digression: The Basel Problem
Aside: Euler used this identity
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
to prove
1 +1
22+
1
32+
1
42+ · · · =
π2
6.
Our goal: prove Wallis’ formula.
Digression: The Basel Problem
Aside: Euler used this identity
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
to prove
1 +1
22+
1
32+
1
42+ · · · =
π2
6.
Our goal: prove Wallis’ formula.
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · ·
(2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2
· 1 · 32 · 2
· 3 · 54 · 4
· 5 · 76 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2
· 3 · 54 · 4
· 5 · 76 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4
· 5 · 76 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6
· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6· · · · · · · · ·
Therefore,
π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
Proof of Wallis’ Formula
sinπx = πx(1− x2)(1− x2
4)(1− x2
9) · · ·
sin(π/2) =π
2·(
3
4
)(15
16
)(35
36
)· · · (2n − 1)(2n + 1)
(2n)2· · ·
1 =π
2· 1 · 3
2 · 2· 3 · 5
4 · 4· 5 · 7
6 · 6· · · · · · · · ·
Therefore,π
2=
2
1· 2
3· 4
3· 4
5· 6
5· 6
7· · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]
sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
An infinite product for cosine
sin(πx) = πx(1− x2)(1− x2
4)(1− x2
9)(1− x2
16)(1− x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− 4x2
4)(1− 4x2
9)(1− 4x2
16)(1− 4x2
25) · · ·
sin(2πx) = 2πx(1− 4x2)(1− x2)(1− 4x2
9)(1− x2
4)(1− 4x2
25) · · ·
sin(2πx) = 2
[πx(1− x2)(1− x2
4) · · ·
] [(1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
]sin(2πx) = 2 sin(πx) cos(πx)
Therefore,
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
The infinite product for√
2
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
cos(π/4) =
(3
4
)(35
36
)(99
100
)· · ·
1√2
=
(1 · 32 · 2
)(5 · 76 · 6
)(9 · 11
10 · 10
)· · ·
Therefore,√
2 =
(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · ·
The infinite product for√
2
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
cos(π/4) =
(3
4
)(35
36
)(99
100
)· · ·
1√2
=
(1 · 32 · 2
)(5 · 76 · 6
)(9 · 11
10 · 10
)· · ·
Therefore,√
2 =
(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · ·
The infinite product for√
2
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
cos(π/4) =
(3
4
)(35
36
)(99
100
)· · ·
1√2
=
(1 · 32 · 2
)(5 · 76 · 6
)(9 · 11
10 · 10
)· · ·
Therefore,√
2 =
(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · ·
The infinite product for√
2
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
cos(π/4) =
(3
4
)(35
36
)(99
100
)· · ·
1√2
=
(1 · 32 · 2
)(5 · 76 · 6
)(9 · 11
10 · 10
)· · ·
Therefore,√
2 =
(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · ·
The infinite product for√
2
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
cos(π/4) =
(3
4
)(35
36
)(99
100
)· · ·
1√2
=
(1 · 32 · 2
)(5 · 76 · 6
)(9 · 11
10 · 10
)· · ·
Therefore,
√2 =
(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · ·
The infinite product for√
2
cos(πx) = (1− 4x2)(1− 4x2
9)(1− 4x2
25) · · ·
cos(π/4) =
(3
4
)(35
36
)(99
100
)· · ·
1√2
=
(1 · 32 · 2
)(5 · 76 · 6
)(9 · 11
10 · 10
)· · ·
Therefore,√
2 =
(2
1· 2
3
)(6
5· 6
7
)(10
9· 10
11
)· · ·
That’s all folks!
THE END
References
Su, Francis E., et al. ”Wallis’ Formula.” Mudd Math FunFacts. http://www.math.hmc.edu/funfacts.
Vandervelde, S., “Newton’s Sums and the Infinite ProductRepresentation for sinπx ,” Mathematics and InformaticsQuarterly, 9 (1999), pp. 64-69.
References
Su, Francis E., et al. ”Wallis’ Formula.” Mudd Math FunFacts. http://www.math.hmc.edu/funfacts.
Vandervelde, S., “Newton’s Sums and the Infinite ProductRepresentation for sinπx ,” Mathematics and InformaticsQuarterly, 9 (1999), pp. 64-69.
