User:Speregrination/Draft

We have $$x=a^{1/3}$$ or $$x^3=a$$.

We take $$f(x)=x^3-a$$. Since$$\xi$$ is the exact root, we have $$\xi^3=a$$.

Substituting $$x_n=\xi+\epsilon_n,x_{n+1}=\xi+\epsilon_{n+1}$$ and $$a=\xi^3$$ in the given method, we obtain on simplification

$$\epsilon_{n+1}=(p+q+r-1)\xi+(p-2q-5r)\epsilon_n+\frac{1}{\xi}(3q+15r)\epsilon_n^2-\frac{1}{\xi^2}(4q+35r)\epsilon_n^3+\cdots$$

For the method to be of order 3, we have

$$p+q+r=1$$,

$$p-2q-5r=0$$,

$$3q+15r=0$$

which gives $$p=5/9,q=5/9,r=-1/9$$.

The error equation becomes $$\epsilon_{n+1}=\frac{5}{3\xi^2}\epsilon_n^3+O(\epsilon_n^4)$$