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I just have a very brief question regarding the formula for error bounds in Newton's method. Depending on where you look, this will either be written as:

$$e_{n+1} \approx \frac{f^{\prime \prime}(r)}{2 f^{\prime}(r)}e_{n}^2$$


$$e_{n+1} \approx -\frac{f^{\prime \prime}(r)}{2 f^{\prime}(r)}e_{n}^2$$

In other words, the sign preceeding the fraction term will differ. Why is this so? I have encountered both versions in different places, and I just wondered if it is arbitrary what you choose to use. After all, if you take the absolute value, the magnitude of the error bound will be the same, but if you leave the answer as it is, without taking the absolute value, then, naturally, one answer will be positive and another will be negative.

If anyone can clear this up for me, I would greatly appreciate it!

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up vote 5 down vote accepted

Let $\alpha$ be the true root. We can define the error $e_{n+1}$ in the estimate $x_{n+1}$ in three different ways.

Way $1$: $\,e_{n+1}=x_{n+1}-\alpha$. If that is the definition, then the first estimate of the error you give is the correct one.

Way $2$: $\,e_{n+1}=\alpha-x_{n+1}$. With that definition, the second estimate you give is the correct one.

Way $3$: $\,e_{n+1}=|\alpha-x_{n+1}|$. Then neither is correct, one should use $e_{n+1} \approx \left|\frac{f^{\prime \prime}(r)}{2 f^{\prime}(r)}\right|e_{n}^2$.

Although one is mostly interested in the absolute value of $\alpha-x_{n+1}$, Ways $1$ and $2$ supply more information than Way $3$, since they tell us whether $x_{n+1}$ is an overestimate or an underestimate.

Each of Ways $1$, $2$, and $3$ is unfortunately in fairly common use.

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Awesome! Thanks a lot for clearing this up. I really appreciate it. – Kristian Aug 9 '12 at 17:29

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