Real root of x3-2x plus 1 in newton raphson method?

Answer 1

By x3 I assume that you mean x3. In which case f(x)=x3-2x+1, and f'(x)=3x2-2.

Therefore our iteration formula is:

xn+1=xn- (xn3-2xn+1)/(3xn2-2)

Starting with x0=0 we get:

x1=0.5

x2=0.6

x3=0.617391304

x4=0.618033095

x5=0.618033988

x6=0.618033988

Starting with x0=0.9 we get:

x1=1.065116279

x2=1.009457333

x3=1.000255451

x4=1.000000195

x5=1

x6=1

Starting with x0=-1.5 we get:

x1=-1.631578947

x2=-1.618183589

x3=-1.618034007

x4=-1.618033989

x5=-1.618033989

The 3 real roots to f(x) are x=-1.618033989, x=0.618033988, and x=1