Want to apply Newtons Method to solve f(x)=0 where f is such that |f ''(x)|≤ 10 and |f ' (x)| ≥ 2 for all x. How do I explain that $x_0$ can take any positive starting value for the method to converge i.e $x_0$ does not need to be specifically close to $τ$ (where $τ$ is the true solution)
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Are you assuming that $f' >0$ and $f''>0$? You mentioned it in the title but then you've used absolute values. – PierreCarre Oct 19 '20 at 09:29
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https://math.stackexchange.com/questions/946400/sufficient-conditions-for-the-convergence-of-newtons-method – Winther Oct 19 '20 at 09:36
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@Winther The answer you link to does not apply here as you have no bound on $|\tau -x_0|$. – PierreCarre Oct 19 '20 at 10:19
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Some thoughts: Since $f'$ cannot change sign, Newton's method will always send the next estimate of the root in the direction of the root. The only possibility for divergence is then that Newton's method will overshoot the root too far. This is likely where $f''$ comes in but I'm not sure. – Simply Beautiful Art Oct 19 '20 at 13:34