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I am interested in using a result about Newton's method, which basically states that if f is convex on $[a,b]$ and it holds $f(a)<0$ and $f(b)>0$, then the Newton iteration converges to $x^*\in[a,b]$ with $f(x^*)=0$ for every starting value $x_0\in[a,b]$ with $f(x_0)\geq 0$.

The proof is not that complicated. Unfortunately, I haven't found this theorem in English literature yet (only in some German lecture notes). Could you recommend me a book or a paper, where this might be found?

Please note: I have also asked this question on math.SE (but without any success).

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    $\begingroup$ here's proof in English: personal.psu.edu/jjb23/web/html/sl455SP12/ch2/13Newton.pdf $\endgroup$
    – Carlo Beenakker
    Commented Nov 14, 2013 at 18:49
  • $\begingroup$ Unfortunately, this is a slightly different result. It applies only to functions which are convex and have one root on $\mathbb{R}$. I can guarantee that there is only one root and convexity on an interval $[a,b]$. But I can't guarantee neither for $\mathbb{R}$ as a whole. $\endgroup$
    – Igor
    Commented Nov 14, 2013 at 18:59

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Global convergence of Newton's method on an interval, Lars Thorlund-Petersen (2004).

Global convergence of Newton's method is considered in the strong sense of convergence for any initial value in an interval. The class of functions for which the method converges globally is characterized. This class contains all increasing convex and increasing concave functions as well as sums of such functions on the given interval.

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  • $\begingroup$ I accepted this answer, because it is a somewhat stronger result (even though it is applied to a slightly modified Newton's Method with an iteration scheme as described by eq. (3) in the paper). It would be also interesting to have a proof, that the newton iteration is monotonically decreasing when starting at $x_0=b$ in the situation above. $\endgroup$
    – Igor
    Commented Nov 20, 2013 at 11:51

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