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有一类计算题,已知函数f(x),求这个函数f(x)在许多点处的函数值之和。对这类计算题,往往直接计算甚是繁杂,费时费力,但根据已知函数f(x)的特点,通过两两配对,而恰恰每一对的和又是定值,这时,可巧妙地解决问题,下面举例说明。例1 (2002年全国高考题)如果函数f(x)=x~2/(1+x~2),则f(1)+f(2)+…+f(9)+f(1/2)+f(1/3)+…+f(1/9)=____。解因为f(x)=x~2/(1+x~2),则f(1/x)=1(1+x~2)(x不为0),恰有f(x)+f(1/x)=1。于是将f(2)与f(1/2)、f(3)与f(1/3)、…、f(9)与f(1/9)配对,得所求的结果为17/2。
There is a type of calculation problem, known as the function f(x), and the sum of the function values of this function f(x) at many points. For such calculation problems, the direct calculation is very complicated and time-consuming. However, according to the characteristics of the known function f(x), pairwise pairing is used, and exactly the sum of each pair is fixed value. In this case, it can be ingenious. To solve the problem, the following examples illustrate. Example 1 (National College Entrance Examination Question in 2002) If the function f(x) = x~2/(1+x~2), then f(1)+f(2)+...+f(9)+f(1/ 2) +f(1/3)+...+f(1/9)=____. Solution because f (x) = x ~ 2 (1 + x ~ 2), then f (1/x) = 1 (1 + x ~ 2) (x is not 0), just f (x) + f (1/x)=1. Then pair f(2) with f(1/2), f(3) and f(1/3), ..., f(9) and f(1/9) to get the result of 17/2 .