Question

设定义在R的函数f(x)同时满足以下条件①f(x)+f(-x)=0 ②f(x)=f(x+2) ③当0≦x<1时f(x)=2^x-1

则f(1/2)+f(1)+f(3/2)+f(2)+f(5/2)=？
要详细步骤

Answer 1

f(5/2)=f(5/2-2)=f(1/2)
f(3/2)=-f(-3/2)=-f(-3/2+2)=-f(1/2)
f(1)=-f(-1)=-f(-1+2)=-f(1),得f(1)=0
f(2)=-f(-2)=-f(-2+2)=-f(0)=-f(2),得f(2)=0
则f(1/2)+f(1)+f(3/2)+f(2)+f(5/2)
=f(1/2)+0-f(1/2)+0+f(1/2)
=f(1/2)
=2^(1/2-1)
=√2/2