有一個國一的數學問題想請教各位大大

2013-10-03 5:49 am
定義在實數上的函數滿足f(0)=0,f(x)+f(1-x)=1,f(x/5)=1/2 f(x),
且當0不大於x_1時,且x_1<x_2,x_2不大於1時
滿足f(x_1)不大於f(x_2),試求f(1/2013)之值?

回答 (1)

2013-10-03 9:24 am
✔ 最佳答案
f(0)=0, f(1)=1
f(1/2)+f(1/2)= 1, then f(1/2)= 1/2
f(1/5)= (1/2) f(1)= 1/2, then f(4/5)= 1/2
while x1< x2, f(x1)<= f(x2) , so f(x)= 1/2 for x=1/5~4/5
then
f(x)= 1/4 for x= 1/25~4/25
f(x)= 1/8 for x= 1/125~4/125
f(x)= 1/16 for x= 1/625~4/625
f(x)= 1/32 for x= 1/3125~4/3125
so, f(1/2013)= 1/32


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