103 Trigonometry Problems

146 103 Trigonometry Problems
implying that
t k+1
t k
= 3 − t2 k
1 − 3t 2 k
=
√ √
3 + tk 3 −
1 − √ tk
·
3t k 1 + √ = u k v k .
3t k
Consequently,
n∏
(u k v k ) = t 2
· t3 ···tn+1
=
t 1 t 2 t n
k=1
establishing equation (∗).
(
tan
tan
π +
3 π n −1
(
π
3 n −1
)
) = 1,
24. [China 1999, by Yuming Huang] Let P 2 (x) = x 2 − 2. Find all sequences of
polynomials {P k (x)} ∞ k=1 such that P k(x) is monic (that is, with leading coefficient
1), has degree k, and P i (P j (x)) = P j (P i (x)) for all positive integers i
and j.
Solution: First, we show that the sequence, if it exists, is unique. In fact, for
each n, there can be only one P n that satisfies P n (P 2 (x)) = P 2 (P n (x)). Let
By assumption,
P n (x) = x n + a n−1 x n−1 +···+a 1 x + a 0 .
(x 2 − 2) n + a n−1 (x 2 − 2) n−1 +···+a 1 (x 2 − 2) + a 0
= (x n + a n−1 x n−1 +···+a 1 x + a 0 ) 2 − 2.
Consider the coefficients on both sides. On the left side, 2i is the highest power
of x in which a i appears. On the right side, the highest power is x n+i , and there
it appears as 2a i x n+i . Thus, we see that the maximal power of a i always is
higher on the right side. It follows that we can solve for each a i in turn, from
n − 1 to 0, by equating coefficients. Furthermore, we are guaranteed that the
polynomial is unique, since the equation we need to solve to find each a i is
linear.
(
Second, we define P n explicitly. We claim that P n (x) = 2T x2
)
n (where Tn is
the nth Chebyshev polynomial defined in Introductory Problem 49). That is,
P n is defined by the recursive relation P 1 (x) = x, P 2 (x) = x 2 − 2, and
P n+1 (x) = xP n (x) − P n−1 (x).