Advanced book on Mathematics Olympiad

Real Analysis 555

which allows us to write the function as an expression with no fractions:

f(x)=(cos 2x+cos 4x+···+cos 2nx)^2 +(sin 2x+sin 4x+···+sin 2nx)^2.

Expanding the squares, we obtain

f(x)=n+

∑

1 ≤l<k≤n

(2 sin 2lxsin 2kx+2 cos 2lxcos 2kx)

=n+ 2

∑

1 ≤l<k≤n

cos 2(k−l)x=n+

n∑− 1

m= 1

2 (n−m)cos 2mx.

In conclusion, the nonzero Fourier coefficients offarea 0 =nanda 2 m= 2 (n−m),
m= 1 , 2 ,...,n−1.
(D. Andrica)

497.Expand the functionfas a Fourier series

f(x)=

∑∞

n= 1

ansinnx,

where

an=

2

π

∫π

0

f(t)sinntdt.

This is possible, for example, sincefcan be extended to an odd function on[−π, π].
Fixn≥2, and consider the functiong:[ 0 ,π]→R,g(x)=nsinx−sinnx. The
functiongis nonnegative because of the inequalityn|sinx|≥|sinnx|,x∈R, which
was proved in the section on induction.
Integrating repeatedly by parts and using the hypothesis, we obtain

(− 1 )m

∫π

0

f(^2 m)(t)sinntdt=n^2 man

π

2

, form≥ 0.

It follows that

(− 1 )m

∫π

0

f(^2 m)(x)(nsinx−sinnx)dx=(na 1 −n^2 man)

π

2

≥ 0.

Indeed, the first term is the integral of a product of two nonnegative functions. This must
hold for any integerm; hencean≤0 for anyn≥2.
On the other hand,f(x)≥0, andf′′(x)≤0 forx∈[ 0 ,π]; hencef(x)−f′′(x)≥ 0
on[ 0 ,π]. Integrating twice by parts, we obtain