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I'm trying to determine why the upper bound for the first approximation $p_0$ of the root $p$ of a function $f \in C^2$ must satisfy the condition: $$|p-p_0|\leq \frac{2|f'(p)|}{|f''(p)|}$$ when $p$ is a simple root, i.e.: $f(p)=0$, but $f'(p)\neq 0$. I've tried to rationalize this from the theorem of convergence of Newton-Raphson's method for a simple root: $$ \frac{|p_{n+1}-p|}{|p_n-p|^2} = \frac{|f''(p)|}{2|f'(p)|}, n \gg 0 $$ , but I suspect I'm moving in the wrong direction. I know the solution has something to do with the second Taylor expansion of the function about $p_0$ when deriving Newton's method (I'm working out of Burden and Faires' Numerical Analysis textbook): $$ f(p) = 0 = f(p_0)+f'(p_0)(p-p_0)+\frac{1}{2!}f''(\xi)(p-p_0)^2 $$, where $\xi$ lies between $p_0$ and $p$. (I know you have to assume that since $p-p_0$ is small, $(p-p_0)^2$ is negligible and therefore you can drop the error term.) Help would be greatly appreciated -- I'm studying for my exams.

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