Find the Root of the equation: $x^3 = x^2 + x + 1$ in the interval [1,3] by Newton's method using double precision. Use $x_0 = 1.5$ and iterate 5 steps. Make a table that shows the number of correct digits in each step.

I think I know what I'm doing, I just need someone to verify that I know what I'm doing.

so I know $f(x) = x^3 - x^2 + x + 1 = 0$ and $f'(x) = 3x^2 - 2x + 1 = 0$

So I know $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$ so I have that $x_1 = 1.5 - \frac{1.5^3 - 1.5^2 + 1.5 + 1}{3(1.5)^2 - 2(1.5) + 1} = 0.7368421053$?

I need to compute $x_2,x_3,x_4,x_5$ but I wanted to make sure I was doing $x_1$ correctly first..

would my table look similar to this? (Sorry didnt know how to make a table in LaTeX lol)


$1$..................$0.7368421053$.........Not Sure What Goes Here


If I may suggest, keep Newton formula in the general way : starting from a guess $x_0$, the method will update it according to $$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$$ So, using $x_0=1.5$, the successive iterates will be $$x_1=+0.7368421053$$ $$x_2=-0.6430644958$$ $$x_3=-0.5516130304$$ $$x_4=-0.5437441177$$ $$x_5=-0.5436890154$$ $$x_6=-0.5436890127$$ which is the solution for ten significant figures.


This answer was for the equation you wrote. However, this is not the equation you need to solve. If $$x^3 = x^2 + x + 1$$ then $$f(x)=x^3-x^2-x-1$$ So, just repeat the above process.

  • $\begingroup$ To start with Newton's method, I would suggest to first try to find square root of a number A, where the polynomial is much simpler. $ x_{n+1}= x_n - \frac{x^2-A}{2 x} $. Tabulate successive roots. Also seeing sketched tangents converging to intersection at x-axis root is instructive. $\endgroup$
    – Narasimham
    Sep 22 '15 at 6:53

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