410 Algebra

(

2 cos

π

5

,2 cos

π

5

,2 cos

3 π

5

,2 cos

3 π

5

,2 cosπ

)

.

(Romanian Mathematical Olympiad, 2002, proposed by T. Andreescu)

197.The Lagrange interpolation formula applied to the Chebyshev polynomialTn− 1 (x)
and to the pointsx 1 ,x 2 ,...,xngives

Tn− 1 (x)=

∑n

k= 1

Tn− 1 (xk)

(x−x 1 )···(x−xk− 1 )(x−xk+ 1 )···(x−xn)

(xk−x 1 )···(xk−xk− 1 )(xk−xk+ 1 )···(xk−xn)

.

Equating the leading coefficients on both sides, we obtain

2 n−^2 =

∑n

k= 1

Tn− 1 (xk)

(xk−x 1 )···(xk−xk− 1 )(xk−xk+ 1 )···(xk−xn)

.

We know that the maximal variation away from 0 ofTn− 1 (x) is 1; in particular,
|Tn− 1 (xk)|≤1,k= 1 , 2 ,...,n. Applying the triangle inequality, we obtain

2 n−^2 ≤

∑n

k= 1

|Tn− 1 (xk)|

|xk−x 1 |···|xk−xk− 1 ||xk−xk+ 1 |···|xk−xn|

≤

∑n

k= 1

1

tk

.

The inequality is proved.
(T. Andreescu, Z. Feng, 103Trigonometry Problems, Birkhäuser, 2004)

198.Let us try to prove the first identity. Viewing both sides of the identity as sequences
inn, we will show that they satisfy the same recurrence relation and the same initial
condition. For the left-hand side the recurrence relation is, of course,

Tn+ 1 (x)

√

1 −x^2

= 2 x

Tn(x)

√

1 −x^2

−

Tn+ 1 (x)

√

1 −x^2

,

and the initial condition isT 1 (x)/

√

1 −x^2 =x/

√

1 −x^2. It is an exercise to check that
the right-hand side satisfies the same initial condition. As for the recurrence relation, we
compute

dn+^1

dxn+^1

( 1 −x^2 )n+^1 −

(^12)

dn
dxn
d
dx
( 1 −x^2 )n+^1 −
(^12)

=

dn

dxn

(

n+ 1 −

1

2

)

( 1 −x^2 )n−

(^12)
(− 2 x)
=−( 2 n+ 1 )x
dn
dxn
( 1 −x^2 )n−
(^12)
−n( 2 n+ 1 )
dn−^1
dxn−^1
( 1 −x^2 )n−
(^12)
.
Here we apply the Leibniz rule for the differentiation of a product to obtain