Richard Belsho Missouri State University MAA Student ...
83
Wallis’ Formula and Other Infinite Products Richard Belshoff Missouri State University MAA Student Chapter Talk January 29, 2009
Transcript of Richard Belsho Missouri State University MAA Student ...
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.
