Question

用对分法求方程x^2+x-1=0的一个正根,精确度为0.01

Answer 1

【解】
令f(x)=x2+x-1
f(0)=-1，f(1)=1，所以取x1=(0+1)/2=1/2
f(x1)=f(½)=-¼，f(1)=1，所以取x2=(½+1)/2=3/4
f(x2)=f(¾)=5/16，f(½)=-¼，所以取x3=(¾+½)=5/8
f(x3)=f(⅝)=1/64，f(½)=-¼，所以取x4=(⅝+½)=9/16
f(x4)=f(9/16)=-31/256，f(⅝)=1/64，所以取x5=(⅝+9/16)=19/32
f(x5)=f(19/32)=-55/1024，f(⅝)=1/64，所以取x6=(⅝+19/32)=39/64
f(x6)=f(39/64)=-79/4096，f(⅝)=1/64，所以取x7=(⅝+39/64)=79/128
f(x7)=f(79/128)=-31/16384，满足|f(x7)-0|<0.01
∴x=79/128 即为满足题目要求的解