Advanced book on Mathematics Olympiad

166 3 Real Analysis

LetCn(x)=

∑∞

n= 1

1

n^2 + 1 cosnxandSn(x)=

∑∞

n= 1

n

n^2 + 1 sinnx. They satisfy

1

2

+Cn(x)+Sn(x)=

πex

e^2 π− 1

,

1

2

+Cn(x)−Sn(x)=

πe−x

1 −e−^2 π

.

Solving this linear system, we obtain

Cn(x)=

1

2

[

πex

e^2 π− 1

+

πe−x

1 −e−^2 π

− 1

]

.

The sum from the statement isC( 1 ). The answer to the problem is therefore

C( 1 )=

1

2

[

πe

e^2 π− 1

+

πe−^1

1 −e−^2 π

− 1

]

. 

We find even more exciting a fundamental result of ergodic theory that proves that
for an irrational numberα, the fractional parts ofnα,n≥1, are uniformly distributed
in[ 0 , 1 ]. For example, whenα=log 10 2, we obtain as a corollary that on average, the
first digit of a power of 2 happens to be 7 as often as it happens to be 1. Do you know a
power of 2 whose first digit is 7?

Theorem.Letf:R→Rbe a continuous function of period 1 and letαbe an irrational
number. Then

lim

n→∞

1

n

(f (α)+f( 2 α)+···+f (nα))=

∫ 1

0

f(x)dx.

Proof.If we approximatefby a trigonometric polynomial with error less than, then
both^1 n(f (α)+f( 2 α)+···+f (nα))and

∫ 1

0 f(x)dxare evaluated with error less than
. Hence it suffices to check the equality term by term for the Fourier series off. For
the constant term the equality is obvious. To check that it holds forf(x)=cos 2πmx
orf(x)=sin 2πmx, withm≥1, combine these two using Euler’s formula into

e^2 πimx=cos 2πmx+isin 2πmx.

We then have

1

n

(

e^2 πimα+e^2 πi^2 mα+···+e^2 π inmα

)

=

e^2 πi(n+^1 )mα− 1

n(e^2 πimα−e^2 πimα)

=

cos 2π(n+ 1 )mα+isin 2π(n+ 1 )mα− 1

n(cos 2πmα+isin 2πmα−cos 2πmα+isin 2πmα)

,

which converges to 0 asn→∞. And for the right-hand side,