PUMaC 2023 · 代数（A 组） · 第 7 题
PUMaC 2023 — Algebra (Division A) — Problem 7
题目详情
- Let S be the set of degree 4 polynomials f with complex number coefficients satisfying 2 3 4 5 f (1) = f (2) = f (3) = f (4) = f (5) = 1. Find the mean of the fifth powers of the constant terms of all the members of S .
解析
- Let S be the set of degree 4 polynomials f with complex number coefficients satisfying 2 3 4 5 f (1) = f (2) = f (3) = f (4) = f (5) = 1. Find the mean of the fifth powers of the constant terms of all the members of S . Proposed by Michael Cheng Answer: 1643751 Let N = 5 for convenience. By the given condition, f ( n ) = ζ for 1 ≤ n ≤ N , where ζ is an n n n -th root of unity. Since f is a degree N − 1 polynomial, the Lagrange interpolation formula 3 P Q N x − m implies that f ( x ) = f ( n ) , where the product runs over m ∈ { 1 , . . . , N } , n =1 m ̸ = n n − m P Q N − m m ̸ = n . We desire the constant term of f , namely f (0) = f ( n ) . Note n =1 m ̸ = n n − m Q ( − 1) ··· ( − ( n − 1)) ( n +1) ··· N N N m n − 1 n − 1 that = · = ( − 1) . Let r := ( − 1) , so that n m ̸ = n n − m ( n − 1) ··· (1) 1 ··· ( N − n ) n n P N f (0) = ζ r . n n n =1 M We now consider f (0) , where M = 5 for convenience. Expand the power to obtain X M α α α α M 1 N 1 N f (0) = ζ · · · ζ · r · · · r · (8) 1 1 N N α | α | = M Here the sum runs over all N -tuples α = ( α , . . . , α ) of nonnegative integers satisfying 1 N P N M M ! α = M , and the multinomial coefficient := counts the number of ways a n n =1 α α ! ··· α ! 1 N given summand occurs. Note that averaging over all possible f is equivalent to averaging over all possible N -tuples ( ζ , . . . , ζ ). Therefore if a given α is such that n does not divide α 1 N n P α n for some 1 ≤ n ≤ N , then ζ = 0 (where the sum runs over all n -th roots of unity ζ ), n n ζ n hence α contributes zero to the average. In other words, the only N -tuples α that contribute to the average are those for which n divides α for all 1 ≤ n ≤ N ; and further the contribution n α α M 1 N of such an α is simply r · · · r · . Call these N -tuples good . We enumerate such good 1 N α N -tuples, using the fact that N = 5 and M = 5. The partitions of M = 5 are: 5, 4 + 1, 3 + 2, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1 + 1, and 1 + 1 + 1 + 1 + 1. Note that for any positive integer d , a good tuple cannot have more than τ ( d ) indices n for which α | d , where τ denotes the number n of divisors of d . Applying this fact to d = 1 and d = 2 eliminates the fourth, fifth, and sixth partitions above. The only valid partitions are 5, 4 + 1, and 3 + 2. The partition 5 can correspond to two good tuples: α with α = 5 and α = 0 for n ̸ = 1; or α 1 n 5 5 5 with α = 5 and α = 0 for n ̸ = 5. By our formula above, these contribute ( r + r ) to the 5 n 1 5 5 average. The partition 4 + 1 can correspond to two good tuples: α with α = 1, α = 4, and α = 0 1 2 n otherwise; or α with α = 1, α = 4, and α = 0 otherwise. By our formula above, these 1 4 n 5 1 4 1 4 contribute ( r r + r r ) to the average. 1 2 1 4 4 The partition 3 + 2 can correspond to three good tuples: α with α = 2, α = 3, and α = 0 1 3 n otherwise; α with α = 3, α = 2, and α = 0 otherwise; or α with α = 2, α = 3, and α = 0 1 2 n 2 3 n 5 2 3 3 2 2 3 otherwise. By our formula above, these contribute ( r r + r r + r r ) to the average. 1 3 1 2 2 3 3 Therefore our answer is 5 5 5 5 5 1 4 1 4 2 3 3 2 2 3 ( r + r ) + ( r r + r r ) + ( r r + r r + r r ) (9) 1 5 1 2 1 4 1 3 1 2 2 3 5 4 3 5 n − 1 where r = ( − 1) implies r = 5, r = − 10, r = 10, r = − 5, and r = 1. Plugging in n 1 2 3 4 5 n yields the answer of 1643751, as desired.