1997 AMC 12 第 27 题

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27.

考虑所有对任意实数 xx 都满足 f(x+4)+f(x4)=f(x)f(x+4)+f(x-4)=f(x) 的函数 ff。每个这样的函数都是周期函数,并且存在一个它们共有的最小正周期 pp。求 pp

Consider those functions ff that satisfy f(x+4)+f(x4)=f(x)f(x+4)+f(x-4)=f(x) for all real x.x. Any such function is periodic, and there is a least common positive period pp for all of them. Find p.p.

88

1212

1616

2424

3232

答案:D
知识点:functional equationsrecurrencesperiodicity
难度评级:2180
小提示:

固定 xx,研究数列 un=f(x+4n)u_n=f(x+4n)

For fixed xx, study the sequence un=f(x+4n)u_n=f(x+4n)

大提示:

反复使用 un+1+un1=unu_{n+1}+u_{n-1}=u_n,找出任意初始二元组何时恢复原状

Use un+1+un1=unu_{n+1}+u_{n-1}=u_n repeatedly to find when every initial pair returns

解答:

un=f(x+4n)u_n=f(x+4n),原方程化为 un+1=unun1u_{n+1}=u_n-u_{n-1}。从 u0,u1u_0,u_1 出发,数列依次为 u0,u1,u1u0,u0,u1,u0u1,u0,u1, \begin{gathered} u_0,u_1,u_1-u_0,-u_0,\\ -u_1,u_0-u_1,u_0,u_1,\ldots \end{gathered}\text{,}因此每个这样的函数都有周期 64=246\cdot4=24。又因为 f(x)=sin(πx12)f(x)=\sin(\frac{\pi x}{12}) 满足原方程且最小正周期为 2424,所以这个公周期不能再小。故正确答案是 D

For un=f(x+4n),u_n=f(x+4n), the equation is un+1=unun1.u_{n+1}=u_n-u_{n-1}. Starting from u0,u1,u_0,u_1, the sequence is u0,u1,u1u0,u0,u1,u0u1,u0,u1,, \begin{gathered} u_0,u_1,u_1-u_0,-u_0,\\ -u_1,u_0-u_1,u_0,u_1,\ldots, \end{gathered} so every such function has period 64=24.6\cdot4=24. This is least because f(x)=sin(πx12)f(x)=\sin(\frac{\pi x}{12}) satisfies the equation and has fundamental period 24.24. Hence D is correct.

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