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有些数学问题,表面上看与周期毫无关系,但实际上隐含着周期性,一旦揭示了周期,问题就迎刃而解。下面以函数和数列为例说明如下。 1．函数中的周期例1 设对任意整数x,都满足f(x)=f(x+1)+f(x-1),且f(0)=19,f(4)=93,求f(59)的值。解∵ f(x)=f(x-1)+f(x+1), ∴ f(x+1)=f(x)+f(x+2), 两式相加并整理得f(x-1)=-f(x+2), ∴ f(x)=-f(x+3), ∴ f(x+6)=-f(x+3)=f(x), 从而f(x)是以6为周期的函数。∴ f(59)=f(6×9+5)=f(5) =f(4)+f(6)=f(4)+f(0)=112。例2 函数f(x)在R上是有定义的,且满足(1)f(x)是偶函数,且f(0)=2008；(2)g(x)=f(x-1)是奇函数。试求f(2004)的值。
Some mathematics problems, on the surface, have nothing to do with the cycle, but they actually imply a periodicity. Once the cycle is revealed, the problem is easily solved. The following is a description of functions and sequences as follows. 1. The cycle example in function 1 is set to any integer x, satisfying f(x)=f(x+1)+f(x-1), and f(0)=19,f(4)=93,finding f The value of (59). Solve f(x)=f(x-1)+f(x+1), ∴ f(x+1)=f(x)+f(x+2), and add two equations to get f( X-1)=-f(x+2), ∴ f(x)=-f(x+3), ∴ f(x+6)=-f(x+3)=f(x), and f (x) is a function of 6 cycles. ∴ f(59)=f(6×9+5)=f(5)=f(4)+f(6)=f(4)+f(0)=112. Example 2 The function f(x) is defined on R and satisfies (1) f(x) is an even function, and f(0)=2008; (2) g(x)=f(x-1) Is an odd function. Try to find the value of f(2004).