HMMT 二月 2011 · ALGCALC 赛 · 第 16 题
HMMT February 2011 — ALGCALC Round — Problem 16
题目详情
- Let f ( x ) = x − r x + r for all real numbers x , where r and r are some real numbers. Define a 2 3 2 3 sequence { g } for all nonnegative integers n by g = 0 and g = f ( g ). Assume that { g } satisfies n 0 n +1 n n the following three conditions: (i) g < g and g > g for all 0 ≤ i ≤ 2011; (ii) there exists 2 i 2 i +1 2 i +1 2 i +2 a positive integer j such that g > g for all i > j , and (iii) { g } is unbounded. If A is the greatest i +1 i n number such that A ≤ | r | for any function f satisfying these properties, find A . 2
解析
- Let f ( x ) = x − r x + r for all real numbers x , where r and r are some real numbers. Define a 2 3 2 3 sequence { g } for all nonnegative integers n by g = 0 and g = f ( g ). Assume that { g } satisfies n 0 n +1 n n the following three conditions: (i) g < g and g > g for all 0 ≤ i ≤ 2011; (ii) there exists 2 i 2 i +1 2 i +1 2 i +2 a positive integer j such that g > g for all i > j , and (iii) { g } is unbounded. If A is the greatest i +1 i n number such that A ≤ | r | for any function f satisfying these properties, find A . 2 Answer: 2 Consider the function f ( x ) − x . By the constraints of the problem, f ( x ) − x must be negative for some x , namely, for x = g , 0 ≤ i ≤ 2011. Since f ( x ) − x is positive for x of large 2 i +1 absolute value, the graph of f ( x ) − x crosses the x -axis twice and f ( x ) − x has two real roots, say a < b . Factoring gives f ( x ) − x = ( x − a )( x − b ), or f ( x ) = ( x − a )( x − b ) + x . Now, for x < a , f ( x ) > x > a , while for x > b , f ( x ) > x > b . Let c 6 = b be the number such that f ( c ) = f ( b ) = b . Note that b is not the vertex as f ( a ) = a < b , so by the symmetry of quadratics, c r r +1 b + c b + a 2 2 exists and = as the vertex of the parabola. By the same token, = is the vertex of 2 2 2 2 f ( x ) − x . Hence c = a − 1. If f ( x ) > b then x < c or x > b . Consider the smallest j such that g > b . j Then by the above observation, g < c . (If g ≥ b then f ( g ) ≥ g ≥ b so by induction, g ≥ g for j − 1 i i i i +1 i all i ≥ j . Hence j > 1; in fact j ≥ 4025.) Since g = f ( g ), the minimum value of f is less than c . j − 1 j − 2 b + a − 1 The minimum value is the value of f evaluated at its vertex, , so 2 ( ) b + a − 1 f < c 2 ( ) ( ) b + a − 1 b + a − 1 b + a − 1 − a − b + < a − 1 2 2 2 2 1 − ( b − a ) b − a + 1
- < 0 4 2 2 3 ( b − a ) b − a < − 4 4 2 2 4 < ( b − a − 1) . 2 Then either b − a − 1 < − 2 or b − a − 1 > 2, but b > a , so the latter must hold and ( b − a ) > 9. 2 Now, the discriminant of f ( x ) − x equals ( b − a ) (the square of the difference of the two roots) and 2 2 ( r + 1) − 4 r (from the coefficients), so ( r + 1) > 9 + 4 r . But r = g > g = 0 so | r | > 2. 2 3 2 3 3 1 0 2 We claim that we can make | r | arbitrarily close to 2, so that the answer is 2. First define G , i ≥ 0 2 i √ 2 as follows. Let N ≥ 2012 be an integer. For ε > 0 let h ( x ) = x − 2 − ε, g ( x ) = − x + 2 + ε and ε G = 2 + ε , and define G recursively by G = g ( G ) , G = h ( G ). (These two equations are 2 N +1 i i ε i +1 i +1 i consistent.) Note the following. √ √ 2 (i) G < G and G > G for 0 ≤ i ≤ N − 1. First note G = − 4 + 2 ε > − 4 + 2 ε + ε = 2 i 2 i +1 2 i +1 2 i +2 2 N − 2 − ε . Let l be the negative solution to h ( x ) = x . Note that − 2 − ε < G < l < 0 since 2 N h ( G ) > 0 > G . Now g ( x ) is defined as long as x ≥ − 2 − ε , and it sends ( − 2 − ε, l ) into ( l, 0) 2 N 2 N ε and ( l, 0) into ( − 2 − ε, l ). It follows that the G , 0 ≤ i ≤ 2 N are well-defined; moreover, G < l i 2 i and G > l for 0 ≤ i ≤ N − 1 by backwards induction on i , so the desired inequalities follow. 2 i +1 2 (ii) G is increasing for i ≥ 2 N + 1. Indeed, if x ≥ 2 + ε , then x − x = x ( x − 1) > 2 + ε so h ( x ) > x . i Hence 2 + ε = G < G < · · · . 2 N +1 2 N +2 (iii) G is unbounded. This follows since h ( x ) − x = x ( x − 2) − 2 − ε is increasing for x > 2 + ε , so G i i increases faster and faster for i ≥ 2 N + 1. 2 2 Now define f ( x ) = h ( x + G ) − G = x + 2 G x + G − G − 2 − ε . Note G = h ( G ) while 0 0 0 0 i +1 i 0 ∞ g = f ( g ) = h ( g + G ) − G , so by induction g = G − G . Since { G } satisfies (i), (ii), and i +1 i i 0 0 i i 0 i i =0 (iii), so does g . i We claim that we can make G arbitrarily close to − 1 by choosing N large enough and ε small enough; 0 this will make r = − 2 G arbitrarily close to 2. Choosing N large corresponds to taking G to be a 2 0 0 larger iterate of 2 + ε under g ( x ). By continuity of this function with respect to x and ε , it suffices to ε Algebra & Calculus Individual Test take ε = 0 and show that (letting g = g ) 0 ( n ) g (2) = g ( · · · g (2) · · · ) → − 1 as n → ∞ . ︸ ︷︷ ︸ n π But note that for 0 ≤ θ ≤ , 2 ( ) ( ) √ θ π θ g ( − 2 cos θ ) = − 2 − 2 cos θ = − 2 sin = 2 cos − . 2 2 2 ( ( )) ( n ) π π n π ( n ) ( n − 1) Hence by induction, g ( − 2 cos θ ) = − 2 cos − + · · · + ( − 1) θ − . Hence g (2) = g ( − 2 cos 0) n 2 4 2 π π π converges to − 2 cos( − + · · · ) = − 2 cos( ) = − 1, as needed. 2 4 3