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.