2.
By the inductive hypothesis the equation P n (x) = 0 has exactly n + 1 distinct real solutions, while the equations
P n (x) = √
2 and P n (x) = − √
2 have two and no distinct real solutions, respectively. Hence, the sets above being
pairwise disjoint, the equation P n+2 (x) = 0 has exactly n + 3 distinct real solutions. Thus we have proved that,
for each n ∈ N , the equation P n (x) = 0 has exactly n + 1 distinct real solutions, so the answer to the question
posed in this problem is 2005.
n
P
Problem 3. Let S n be the set of all sums
x k , where n ≥ 2, 0 ≤ x 1 , x 2 , . . . , x n ≤ π 2 and
k=1
X
sin x k = 1 .
a) Show that S n is an interval. [10 points]
b) Let l n be the length of S n . Find lim
n→∞ l n . [10 points]Solution. (a) Equivalently, we consider the set
Y = {y = (y 1 , y 2 , ..., y n )| 0 ≤ y 1 , y 2 , ..., y n ≤ 1, y 1 + y 2 + ... + y n = 1} ⊂ R n
and the image f (Y ) of Y under
f (y) = arcsin y 1 + arcsin y 2 + ... + arcsin y n .
Note that f (Y ) = S n . Since Y is a connected subspace of R n and f is a continuous function, the image f(Y ) is
also connected, and we know that the only connected subspaces of R are intervals. Thus S n is an interval.
(b) We prove that
n arcsin 1
n ≤ x 1 + x 2 + ... + x n ≤ π
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