Question

一道函数周期性问题（只需做第三问）谢谢啦

Answer 1

函数周期是2,所以an=f(1/2n)

f(1)=a=f(1/2+1/2)=f(1/2)^2=>f(1/2)=√a,

f(1/2)=f(1/4+1/4)=f(1/4)^2=>f(1/4)=4^√a,

……

f(1/2n)=2^√(f(1/(2n-2))=4^√(f(1/(2n-4))=……=2n^√f(1)=2n^√a