HMMT 二月 2023 · 冲刺赛 · 第 33 题
HMMT February 2023 — Guts Round — Problem 33
题目详情
- [25] Given a function f , let π ( f ) = f ◦ f ◦ f ◦ f ◦ f . The attached sheet has the graphs of ten smooth functions from the interval (0 , 1) to itself. The left-hand side consists of five functions: 1 1 1 1 1 • F ( x ) = 0 . 005 + sin 2 x + sin 4 x + sin 8 x + sin 16 x + sin 32 x ; 1 2 4 8 16 32 • F ( x ) = F ( F ( x + 0 . 25)); 2 1 1 2 • F ( x ) = F ((1 − x ) F ((1 − x ) )); 3 1 1 • F ( x ) = F ( x ) + 0 . 05 sin(2 πx ); 4 1 • F ( x ) = F ( x + 1 . 45) + 0 . 65. 5 1 The right-hand side consists of the five functions A , B , C , D , and E , which are π ( F ), . . . , π ( F ) in some 1 5 order. Compute which of the functions { A, B, C, D, E } correspond to π ( F ) for k = 1, 2, 3, 4, 5. k Your answer should be a five-character string containing A , B , C , D , E , or X for blank. For instance, if you think π ( F ) = A and π ( F ) = E , then you would answer AXXXE . If you attempt to identify n 1 5 2 functions and get them all correct, then you will receive n points. Otherwise, you will receive 0 points.
解析
- [25] Given a function f , let π ( f ) = f ◦ f ◦ f ◦ f ◦ f . The attached sheet has the graphs of ten smooth functions from the interval (0 , 1) to itself. The left-hand side consists of five functions: 1 1 1 1 1 • F ( x ) = 0 . 005 + sin 2 x + sin 4 x + sin 8 x + sin 16 x + sin 32 x ; 1 2 4 8 16 32 • F ( x ) = F ( F ( x + 0 . 25)); 2 1 1 2 • F ( x ) = F ((1 − x ) F ((1 − x ) )); 3 1 1 • F ( x ) = F ( x ) + 0 . 05 sin(2 πx ); 4 1 • F ( x ) = F ( x + 1 . 45) + 0 . 65. 5 1 The right-hand side consists of the five functions A , B , C , D , and E , which are π ( F ), . . . , π ( F ) in 1 5 some order. Compute which of the functions { A, B, C, D, E } correspond to π ( F ) for k = 1, 2, 3, 4, 5. k Your answer should be a five-character string containing A , B , C , D , E , or X for blank. For instance, if you think π ( F ) = A and π ( F ) = E , then you would answer AXXXE . If you attempt to identify 1 5 2 n functions and get them all correct, then you will receive n points. Otherwise, you will receive 0 points. Proposed by: Sean Li F . A . 1 B . F . 2 C . F . 3 D . F . 4 E . F . 5 Answer: DACBE Solution: First, note that F and E are the only functions whose range is outside the range (0 , 0 . 7), 5 so they must be matched. 1 1 2 For graphs F and F , split the square [0 , 1] into a three by three grid of × squares. The only 2 3 3 3 1 2 2 1 squares that the graphs pass through are [ , ) × I and possibly [ , 1] × [0 , ), so after five iterates the 3 3 3 3 1 2 range of these graphs stabilizes to f ([ , )). Graph F corresponds to the flatter graph C , while graph 3 3 3 F to the bumpier graph A . This is good enough for 9 points. 2 It remains to distinguish between F and F . The details are a bit messy, so here are two ideas: 1 4 1 1 • Split the grid in a 5 × 5 grid of -by- squares. It is clear these will distinguish F and F since 1 4 5 5 B does not pass through [0 . 2 , 0 . 4) × [0 . 4 , 0 . 6) but D does not. • Observing the definitions of F and F , we have that the phases of the waves that make up F 1 4 1 “agree” much more than they do in F , since π is irrational. This suggests that π ( F ) will have 4 4 more chaotic behavior than π ( F ). 1 The final answer is DACBE . Remark. The graphs were made using gnuplot .