- #1
peterianstaker
- 2
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Can someone check this is correct?
Using the Newton Raphson method with X0=2 to find the root of the equation:
x^3-x-1=0 (correct to 4.d.p)
My answer is:
f'(x)= 3x^2-1
xn+1= 2-x^3-x-1/3x^2-1
xn+1= 2-2^3-2-1/3(2^2)-1
x1= 17/11
x2= 17/11-(17/11^3)-17/11-1/3x(17/11^2)-1
= 1.3596
Using the Newton Raphson method with X0=2 to find the root of the equation:
x^3-x-1=0 (correct to 4.d.p)
My answer is:
f'(x)= 3x^2-1
xn+1= 2-x^3-x-1/3x^2-1
xn+1= 2-2^3-2-1/3(2^2)-1
x1= 17/11
x2= 17/11-(17/11^3)-17/11-1/3x(17/11^2)-1
= 1.3596