如题所述
第1个回答 2018-09-21
f(x+½)=½+√[f(x)-f²(x)]
f(x+½)-½=√[f(x)-f²(x)]
f²(x+½)- f(x+½)+¼=f(x)-f²(x)
f²(x+½+½)- f(x+½+½)+¼=f(x+½)-f²(x+½)
即f²(x+1)-f(x+1)+¼=-[f²(x+½)- f(x+½)]=¼+f²(x)-f(x)
f²(x+1)-f(x+1)=f²(x)-f(x)
[f(x+1)+f(x)][f(x+1)-f(x)]-[f(x+1)-f(x)]=0
[f(x+1)-f(x)][f(x+1)+f(x)-1]=0
f(x+1)=f(x)
T=1
第2个回答 2018-09-21
f(x) =x(x+1)(x+2)...(x+2n) f'(x) =(x+1)(x+2)...(x+2n)+x(x+2)...(x+2n)+x(x+1)(x+3)...(x+2n)+... +x(x+1)(x+2)...(x+2n-1) f'(-n)=-n(-n+1)(-n+2)....(-n+n-1)(-n+n+1)....(-n+2n) = (-1)^n . (n!)^2本回答被网友采纳
