Exercises 35
5. (Euler) In courses in complex analysis, it is often proved that sin x pos-
sesses the following Weierstrass factorization (valid for all x C):
(1.15) sin x = x

n=1
1
x2
n2π2
;
see, e.g., [Pri01] for a short, direct proof of this identity. A proof using
only real-variable methods appears in [Kob84, Chapter II].
(a) Starting from (1.15), show that
x cot x = 1 2

m=1
ζ(2m)
x2m
π2m
,
where ζ denotes the Euler-Riemann zeta function. Hint: Take the
logarithmic derivative of both sides.
(b) Computing by hand the first few coefficients in the Taylor series for
x cot x about x = 0, check that ζ(2) =
π2/6
and ζ(4) =
π4/90.
6. (J. D. Dixon) We outline Dixon’s proof [Dix62] that π is not the root
of a polynomial over Z of degree 2. The method is that employed
by Niven to show π is irrational (see [Niv47]). Suppose for the sake of
contradiction that π is a root of P (T) = aT
2
+ bT + c, where a, b and c
are integers, not all vanishing.
Given a polynomial f(T) R[T], define
(1.16) F (T) := f(T) f
(2)(T)
+ f
(4)(T)
f
(6)(T)
+ · · · .
Then F (T) R[T]. View F as a function of a real variable x.
(a) Check that
d
dx
(
F (x) sin x F (x) cos x
)
= f(x) sin(x),
and conclude that
(1.17)
π
0
f(x) sin x dx = F (π) + F (0).
(b) With n a positive integer to be chosen shortly, let f be the polyno-
mial
f(T) :=
1
n!
P
(T)2n(P
(T) P
(0))2n.
Show that the left-hand side of (1.17) is strictly between 0 and 1 if
n is sufficiently large.
We now fix such an n and derive a contradiction by showing that
the right-hand side of (1.17) is an integer.
(c) Show that f
(r)(0)
= f
(r)(π)
= 0 for all 0 r 2n.
Previous Page Next Page