(August 24, 2013)
01b. Product expansion of sin x
Paul Garrett [email protected]
http://www.math.umn.edu/egarrett/
[This document is http://www.math.umn.edu/˜garrett/m/mfms/notes 2013-14/01b product expn sine.pdf]
Euler did eventually prove the product expansion
sin πx = πx ·
Y
1−
n≥1
x2 n2
The question does not mention complex numbers at all, but the simplest verification of this product expansion
is a standard application of basic complex analysis, at the level of Liouville’s theorem and Laurent expansions
near poles, providing yet another powerful motivation for understanding basic complex analysis. [1]
[0.1] Partial fraction expansion of
π2
sin2 πx
We claim that there is a partial fraction expansion
X
π2
1
=
2
(z − n)2
sin πz
n∈Z
To see this, first note that both sides have double poles exactly at integers. The Laurent expansion of the
right-hand side near n ∈ Z begins
1
+ holomorphic
(z − n)2
The left-hand side is periodic, so it suffices to see the Laurent expansion near 0:
π2
=
sin2 πz
π2
πz +
(πz)3
3!
+ ...
2 =
1
·
z2 1 −
1
π2 z2
3!
+ ...
2 =
2
π2 z2
1 1
·
1
+
+
.
.
.
= 2 + holomorphic
2
z
3!
z
This Laurent expansion matches that of the partial fraction expansion. Thus,
f (z) =
X
π2
1
−
2
(z − n)2
sin πz
n∈Z
has no poles in C, so is entire. On the real line, after cancellation of poles, f continuous. The periodicity
f (z + 1) = f (z) is visible, so f is bounded on the real line. In fact, since f is bounded on any region
{x + iy : 0 ≤ x ≤ 1, |y| ≤ N }, the periodicity gives the boundedness of f (z) on every band |y| ≤ N
containing R.
Both parts of f (z) go to 0 as |y| → ∞. Thus, f is bounded and entire, so constant, by Liouville’s theorem.
Since f (z) → 0 as |y| → ∞, this constant is 0, giving the desired equality.
[0.2] Partial fraction expansion of π cot πx
Next, regroup slightly, to improve convergence:
X
1
1
1
π2
=
+
+
z2
(z − n)2
(z + n)2
sin2 πz
n≥1
[1] Weierstraß and Hadamard product expansions apply to general entire functions, but with more overhead than
needed here.
1
Paul Garrett: 01b. Product expansion of sin x (August 24, 2013)
The left-hand side is the derivative of −π cot πz, and with the improved convergence the right-hand side is
the obvious termwise derivative, so up to a constant C,
π cot πz = C +
1 X 1
1 +
+
z
z−n z+n
n≥1
The identity
1
2z
1
(z + n) + (z − n)
= 2
+
=
2
2
z−n z+n
z −n
z − n2
certifies that convergence is uniform and absolute, and that the summands are odd functions of z. Everything
but the constant C is odd as a function of z, so C = 0. Thus,
π cot πz =
1 X 1
1 +
+
z
z−n z+n
n≥1
[0.3] Product expansion of sin πx
Also, π cot πz is the logarithmic derivative of sin πz:
d
(sin πz)0
π cos πz
log(sin πz) =
=
= π cot πz
dz
sin πz
sin πz
Thus,
d
(sin πz)0
1 X 1
1 log(sin πz) =
=
+
+
dz
sin πz
z
z−n z+n
n≥1
We intend to integrate. First, anticipating our goal, note that
− n1
d
z
1
log 1 −
=
=
dz
n
1 − nz
z−n
Thus, integrating, for some constant C,
log(sin πz) = C + log z +
X
n≥1
log 1 −
X
z z
z2 + log 1 −
= C + log z +
log 1 − 2
n
n
n
n≥1
Exponentiating,
sin πz = eC · z ·
Y
1−
n≥1
z2 n2
Looking at the power series at z = 0, we see that eC = π, so
Y
sin πz = πz ·
n≥1
2
1−
z2 n2