定义域［0,1］,f(0)=0,f(x)+f(1-x)=1,f(1/5x)=1/2f(x),且当0≤x1≤x2≤1时,f(x1)≤f(x2),求f(1/2008)

Question

定义域［0,1］,f(0)=0,f(x)+f(1-x)=1,f(1/5x)=1/2f(x),且当0≤x1≤x2≤1时,f(x1)≤f(x2),求f(1/2008)

Answer 1

f(0)=0,f(x)+f(1-x)=1 ==> f(1/2)=1/2,f(1)=1
f(1)=1,f(x/5)=1/2f(x),f(x)+f(1-x)=1 ==> f(1/5)=1/2,f(4/5)=1/2
f(1/5)=1/2,f(1/2)=1/2,f(4/5)=1/2,f(x1)≤f(x2){0≤x1 [1/5,4/5]区间f(x)=1/2
f(x/5)=1/2f(x) ==> [1/5^n,4/5^n]区间上f(x)=1/2^n
1/2020 属于 [1/5^5,4/5^5] ==> f(1/2008) = 1/2^5 =1/32