Advanced book on Mathematics Olympiad

Real Analysis 505

whose limit astgoes to infinity is 1. The squeezing principle implies that

tlim→∞

f (mt)

f(t)

= 1 ,

as desired.

(V. Radu)

385.The sum under discussion is the derivative offat 0. We have

∣

∣

∣∣

∣

∑n

k= 1

kak

∣

∣

∣∣

∣

=|f′( 0 )|=lim

x→ 0

∣∣

∣∣f(x)−f(^0 )

x− 0

∣∣

∣∣

=lim

x→ 0

∣∣

∣

∣

f(x)

x

∣∣

∣

∣=xlim→ 0

∣∣

∣

∣

f(x)

sinx

∣∣

∣

∣·

∣∣

∣

∣

sinx

x

∣∣

∣

∣≤^1.

The inequality is proved.

(28th W.L. Putnam Mathematics Competition, 1967)

386.The condition from the statement implies thatf(x)=f(−x), so it suffices to check

thatfis constant on[ 0 ,∞). Forx≥0, define the recursive sequence(xn)≥ 0 ,byx 0 =x,

andxn+ 1 =

√

xn, forn≥0. Then

f(x 0 )=f(x 1 )=f(x 2 )= ··· =f(lim

n→∞

xn).

And limn→∞xn=1ifx>0. It follows thatfis constant and the problem is solved.

387.The answer is yes, there is a tooth function with this property. We constructfto

have local maxima at 22 n^1 + 1 and local minima at 0 and 212 n,n≥0. The values of the

function at the extrema are chosen to bef( 0 )=f( 1 )=0,f(^12 )=^12 , andf( 22 n^1 + 1 )= 21 n

andf( 212 n)= 2 n^1 + 1 forn≥1. These are connected through segments. The graph from

Figure 66 convinces the reader thatfhas the desired properties.

(K ̋ozépiskolai Matematikai Lapok(Mathematics Gazette for High Schools, Bu-

dapest))

388.We prove by induction onnthatf( 3 mn)=0 for all integersn≥0 and all integers

0 ≤m≤ 3 n. The given conditions show that this is true forn=0. Assuming that it is

true forn− 1 ≥0, we prove it forn.

Ifm≡ 0 (mod 3), then

f

(m

3 n

)

=f

( m

3

3 n−^1

)

= 0

by the induction hypothesis.

Ifm≡ 1 (mod 3), then 1≤m≤ 3 n−2 and