m2pm3infprod.tex
III.9. Infinite products for sin, cos and tan.
From III.8 (Lecture 32(11)):
cosec z =
∞
∑
∑
∑
1
(−)n
1
+ 2z
−2z
.
= + 2z
2
2
2
z
z
even
n=1 z − n π
odd
Similarly,
cot z =
∞
∑
1
1
+ 2z
.
2
2 2
z
n=1 z − n π
(i)
(ii)
Now (with D := d/dz the differentiation operator)
1
D log tan z = cosec z,
2
D log(1 −
D log sin z = cot z,
z2
−2z/n2 π 2
2z
)
=
= 2
.
2
2
2
2
2
nπ
1 − z /n π
z − n2 π 2
Integrating (ii) gives (using Π for product, as we do
log sin z − log z =
∞
∑
log(1 −
1
∑
for sum)
z2
z2
∞
)
=
log
Π
(1
−
).
1
n2 π 2
n2 π 2
Taking exponentials,
sin z
z2
= Π∞
(1
−
)
(iii)
1
z
n2 π 2
(both sides → 1 as z → 0, accounting for the constant of integration). Similarly, integrating (i) gives
∑
∑
1
z2
z2
log tan z = log z + log c +
log(1 − 2 2 ).
log(1 − 2 2 ) −
2
nπ
nπ
even
odd
Take exponentials:
1
z2
z2
tan z = czΠeven (1 − 2 2 )/Πodd (1 − 2 2 ).
2
nπ
nπ
Both products → 1 as z → 0, so for small z, LHS ∼ 21 z, RHS ∼ cz: c = 1/2.
Replace z by 2z:
tan z =
z2
4z 2
sin z
∞
(1
−
(1
−
= zΠ∞
)/Π
)
n=1
n=1
cos z
n2 π 2
(2n − 1)2 π 2
1
(iv)
(cancelling 4 in (2z)2 /(2n)2 ). From (iii) and (iv),
4z 2
cosz =
−
).
(v)
(2n − 1)2 π 2
Note that the infinite products for sin and cos display zeros at the integers
and the half-integers, as they should.
Taking z = π/2 in the product (iii) for sin:
1
1
π −1 = Π∞
).
1 (1 −
2
4n2
This is Wallis’ product for π (John WALLIS (1616-1703), Arithmetica infinitorum, 1656 – see Coursework 2 Q5 for Wallis’ product by Real Analysis).
By (iii) and the power series for sin,
Π∞
1 (1
∞
∑
(−)k z 2k
z2
sin z
=
= Π∞
(1
−
).
1
z
n2 π 2
k=0 (2k + 1)!
Equate coefficients of z 2 :
∞
1
1 ∑
1
− = − 2.
:
3!
π 1 n2
ζ(2) =
∞
∑
1/n2 = π 2 /6,
1
4
again. Similarly, equating coefficients of z gives
1
1
1
1
=
= 4 .ΣΣ1≤r<s<∞ 2 2 .
5!
120
π
r s
But
∞
∞
∞
∑
∑
1 ∑
1
1
1
(
).(
)
=
+ 2ΣΣ1≤r<s<∞ 2 2 .
2
2
4
r s
r=1 r
s=1 s
n=1 n
The LHS is ζ(2)2 = (π 2 /6)2 = π 4 /36. By above, the RHS is
2.π 4 /120. So
ζ(4) :=
∞
∑
1
n=1
n4
= π4(
∑∞
1
1/n4 +
1
1
π4
− )=
(10 − 6) = 4π 4 /360 = π 4 /90,
36 60
360
∑
6
again. The same method shows that ζ(6) := ∞
1 1/n is a rational multiple
of π 6 , etc.
We quote (Weierstrass’ product for the Gamma function)
z −z/n
)e
}
1/Γ(z) = zeγz Π∞
n=1 {(1 +
n
(where γ is Euler’s constant – this shows again that Γ has poles, 1/Γ has
zeros, at 0, −1, −2, . . . ). From this and Γ(z)Γ(1 − z) = π/ sin πz, we can
recover the product (iii) for the sin (exercise).
2