2.2 Polynomials 59
Chebyshev’s theorem.For fixedn≥ 1 , the polynomial 2 −n+^1 Tn(x)is the unique monic
nth-degree polynomial satisfying
max
− 1 ≤x≤ 1
| 2 −n+^1 T(x)|≤max
− 1 ≤x≤ 1
|P(x)|,
for any other monicnth-degree polynomialP(x).
One says that among all monicnth-degree polynomials, 2−n+^1 Tn(x)has the smallest
variation away from zero on[− 1 , 1 ]. This variation is 2 n^1 − 1. Let us see how Chebyshev’s
theorem applies to a problem fromChallenging Mathematical Problems with Elementary
Solutionsby A.M. Yaglom and I.M. Yaglom.
Example.LetA 1 ,A 2 ,...,Anbe points in the plane. Prove that on any segment of length
lthere is a pointMsuch that
MA 1 ·MA 2 ···MAn≥ 2
(
l
4
)n
.
Solution.Rescaling, we can assume thatl=2. Associate complex coordinates to points
in such a way that the segment coincides with the interval[− 1 , 1 ]. Then
MA 1 ·MA 2 ···MAn=|z−z 1 |·|z−z 2 |···|z−zn|=|P(z)|,
whereP(z)is a monic polynomial with complex coefficients, andz∈[− 1 , 1 ]. Write
P(z)=R(z)+iQ(z), whereR(z)is the real part andQ(z)is the imaginary part of the
polynomial. Sincezis real, we have|P(z)|≥|R(z)|. The polynomialR(z)is monic,
so on the interval[− 1 , 1 ]it varies away from zero at least as much as the Chebyshev
polynomial. Thus we can findzin this interval such that|R(z)|≥ 2 n^1 − 1. This implies
|P(z)|≥ 2 · 21 n, and rescaling back we deduce the existence in the general case of a point
Msatisfying the inequality from the statement.
Stepping aside from the classical picture, let us also consider the families of polyno-
mialsTn(x)andUn(x)defined byT 0 (x)=2,T 1 (x)=x,Tn+ 1 (x)=xTn(x)−Tn− 1 (x),
andU 0 (x)=1,U 1 (x)=x,Un+ 1 (x) =xUn(x)−Un− 1 (x). These polynomials are
determined by the equalities
Tn
(
z+
1
z
)
=zn+
1
zn
and Un
(
z+
1
z
)
=
(
zn+^1 −
1
zn+^1
)/(
z−
1
z
)
.
Also,Tn(x)=^12 Tn( 2 x)andUn(x)=Un( 2 x). Here is a quickie that usesTn(x).
Example.Letabe a real number such thata+a−^1 is an integer. Prove that for any
n≥1, the numberan+a−nis an integer.