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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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