Proof: Newton-Raphsons numerical method is of second order

Asked Dec 11 '17 at 12:36

Active Dec 11 '17 at 12:36

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In the proof of the fact that Newton-Raphson is a second order numerical method they use a Taylor-sequence.

Iterative formula for Newton-Raphson:

$x_n=x_{n-1} - \frac{F(x_{n-1})}{F'(x_{n-1})}$

Why is there a factor $F''(\xi_{n-1})$ with $\xi_{n-1}$ between $\alpha$ and $x_{n-1}$ in the third term?

asked Dec 11 '17 at 12:36

WinstonCherf

If you work with a Taylor approximation, you should also work with some form of error term. – Hagen von Eitzen Dec 11 '17 at 12:37
Ok, but how do you know the error term is in factor nr. 3? And Why isn’t there a term $F”(x_{n-1})$ in factor nr. 3? – WinstonCherf Dec 11 '17 at 13:03
This is the Lagrange remainder of a first order Taylor expansion. – Ian Dec 11 '17 at 13:37

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