HMMT 二月 2011 · CALCCOMB 赛 · 第 8 题

HMMT February 2011 — CALCCOMB Round — Problem 8

Let f : [0 , 1) → R be a function that satisfies the following condition: if ∞ ∑ a n x = = .a a a . . . 1 2 3 n 10 n =1 is the decimal expansion of x and there does not exist a positive integer k such that a = 9 for all n n ≥ k , then ∞ ∑ a n f ( x ) = . 2 n 10 n =1 ( ) 1 ′ Determine f . 3

解析

Let f : [0 , 1) → R be a function that satisfies the following condition: if ∞ ∑ a n x = = .a a a . . . 1 2 3 n 10 n =1 is the decimal expansion of x and there does not exist a positive integer k such that a = 9 for all n n ≥ k , then ∞ ∑ a n f ( x ) = . 2 n 10 n =1 ( ) 1 ′ Determine f . 3 ∑ ∞ 1 3 Answer: 0 Note that = . n n =1 3 10 Clearly f is an increasing function. Also for any integer n ≥ 1, we see from decimal expansions that 1 1 1 1 f ( ± ) − f ( ) = ± . n 2 n 3 10 3 10 1 − n − 1 − n Consider h such that 10 ≤ | h | < 10 . The two properties of f outlined above show that | f ( + 3 ∣ ∣ 1 1 ∣ f ( + h ) − f ( ) ∣ 1 1 1 1 n +1 3 3 h ) − f ( ) | < . And from | | ≤ 10 , we get < . Taking n → ∞ gives h → 0 ∣ ∣ 2 n n − 1 3 10 h h 10 ′ 1 1 and f ( ) = lim = 0. n − 1 n →∞ 3 10