MHB Glad to hear that the answer is correct! You are welcome, happy to help.

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The convergence criterion |f(x0)f''(x0)|<|f'(x0)|^2 is confirmed to apply to the Newton-Raphson method, which was initially questioned. The formula is not commonly found in standard resources, leading to some confusion among participants. The discussion includes a reminder of the Newton iterations and the conditions for convergence. Ultimately, the correct answer is affirmed as option (a), the Newton-Raphson method. This clarification helps solidify understanding of the convergence criteria in numerical methods.
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|f(x0)f''(x0)|<|f'(x0)|^2 where I is the interval containing the approximate root x0, is the convergence criterion of ...

(a) Newton - Raphson method
(b) Iteration method
(c) Secant method
(d) False position method

According to me its (a), but I confused because this formula is not directly given anywhere I had searched a lot on the internet, can anybody confirm what is the correct answer

Thanks a lot in advance
 
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ssss said:
|f(x0)f''(x0)|<|f'(x0)|^2 where I is the interval containing the approximate root x0, is the convergence criterion of ...

(a) Newton - Raphson method
(b) Iteration method
(c) Secant method
(d) False position method

According to me its (a), but I confused because this formula is not directly given anywhere I had searched a lot on the internet, can anybody confirm what is the correct answer

Thanks a lot in advance

Wellcome on MHB $SS^{2}$!...

... remembering the Newton iterations...

$$x_{n+1}= x_{n} - \frac{f(x_{n})}{f^{\ '} (x_{n})}\ (1) $$

... if is... $$\frac{|f(x_{0})\ f^{\ '}(x_{0})|}{|f^{\ '\ 2}(x_{0})|}= \frac{|f(x_{0})|}{|f^{\ '}(x_{0}|}= a < 1\ (2)$$

... and that holds for any $x_{0}$. then the sequence of $x_{n}$ obeys to the difference equation... $$ x_{n+1} - x_{n}= (\pm a)^{n}\ (3)$$... and it converges... Kind regards $\chi$ $\sigma$
 
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chisigma said:
Wellcome on MHB $SS^{2}$!...

... remembering the Newton iterations...

$$x_{n+1}= x_{n} - \frac{f(x_{n})}{f^{\ '} (x_{n})}\ (1) $$

... if is... $$\frac{|f(x_{0})\ f^{\ '}(x_{0})|}{|f^{\ '\ 2}(x_{0})|}= \frac{|f(x_{0})|}{|f^{\ '}(x_{0}|}= a < 1\ (2)$$

... and that holds for any $x_{0}$. then the sequence of $x_{n}$ obeys to the difference equation... $$ x_{n+1}= \pm a\ x_{n}\ (3)$$... and it converges... Kind regards $\chi$ $\sigma$

so I was right answer is Newton Rapshon option a, Thanks a lot for explanation
 
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