Table of Contents

Error Estimation and Error Verification of Newton's Method
Let $f$ be a function that satisfies the conditions to apply Newton's Method and converges to the root $\alpha$. We will now develop a way to estimate the error of the approximation iterates $x_n$ (obtained from Newton's Method) from the root $\alpha$. Note that since $\alpha$ is a root of $f$, then $f(\alpha) = 0$ and thus by the Mean Value Theorem, for some $\xi_n$ between $x_n$ and $\alpha$, we have that:
(1)The exact error is given above, however, we don't necessary know what $\xi_n$ is. Instead, let's find a suitable value to replace $\xi_n$ with. Suppose that $x_n$ is close to $\alpha$. Then $\xi_n$ is close to $x_n$ since $\xi_n$ is between $x_n$ and $\alpha$ and thus the error is approximated as:
(2)Now recall that the general iteration formula for Newton's Method is $x_{n+1} = x_n  \frac{f(x_n)}{f'(x_n)}$ and so $\frac{f(x_n)}{f'(x_n)} = x_{n+1}  x_n$. Substituting this into our error approximation above and we get that:
(3)Therefore if we are given an allowable error of $\epsilon$, then if we can ensure that $x_{n+1}  x_n < \epsilon$ then it's likely that the error between $x_n$ and $\alpha$ is less than $\epsilon$ or almost less than $\epsilon$.
Error Verification
We saw above that the error between the root $\alpha$ and our approximation $x$ is approximately equal to $x_{n+1}  x_n$ however, if $x_{n+1}  x_n < \epsilon$ then it is possible that $\alpha  x_n > \epsilon$ still.
To verify that $x_{n+1}  x_n < \epsilon$, we can test to see whether $f(x_n  \epsilon) \cdot f(x_n + \epsilon) < 0$. If so, then we know that a root $\alpha$ exists between them and i within $\epsilon$ of our approximation $x_n$.