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Consider a variation of Newton's method in which only one derivative is needed; that is, $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. Find $C$ and $s$ such that $e_{n+1} = C e^{s}_{n}$.

First of all, I thouhght the above iterative process was the usual variation of newtons method.

Please note my edit at the very bottom

So $e_n = x_n - r$ which means $e_{n+1} = x_n - \frac{f(x_n)}{ f'(x_n)} - r= e_n - \frac{f(x_n)}{ f'(x_n)}$. Eventually by applying Taylor's Theorem, I get $e_{n+1} = C e^{2}_{n}$.

So $s=2$, but how do I find $C$? Isn't it just a constant? I'm a little confused by this question. In more detail $e_{n+1} = \frac{1}{2} \frac{f''(z_n)}{f'(x_n)} e^{2}_{n}\approx \frac{1}{2}\frac{f''(r)}{f'(r)}e^{2}_{n}$. Maybe they just want me to say $C$ is the bit next to the $e^{2}_{n}$.

Can anyone confirm my answer?

It's supposed to be $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_0)}$ unfortunately.

Ozera
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  • @Moo You're exactly correct. I misread. – Ozera Sep 29 '16 at 11:51
  • @Moo How does it being $f'(x_0)$ rather than $f'(x_n)$ change my answer? I'm looking through my work with the taylor series and i'm not sure what would change – Ozera Sep 29 '16 at 11:53
  • @Moo I would still have $e_{n+1} = e_n - \frac{f(x_n)}{f'(x_0)} = \frac{e_n f'(x_0) - f(x_n)}{f'(x_0)}$. $f(r) = f(x_n - e_n) = f(x_n) - e_n f'(x_n) + \frac{1}{2}e^{2}_{n}f''(z_n)$. But now $x_0$ isn't used anywhere. Where is my mistake? – Ozera Sep 29 '16 at 11:59
  • if you replace $x_n$ by $x_0$, it won't be Newton-Raphson's method. – hamam_Abdallah Sep 29 '16 at 12:49
  • @AbdallahHammam The exercise says it is a "variation" of the method. I suppose it is simply for exercise sake. – Ozera Sep 29 '16 at 12:54
  • @Moo In the answers of my guide book it comes out that $C=1-f''(\xi_n)/f'(x_0), s=1$, could you explain to me why? – user482152 Mar 08 '18 at 02:12

1 Answers1

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$e_{n+1}=g(x_{n})=g(e_n+r)=g(r)+e_ng'(r)+C\frac{1}{2}e_n^2$

with $C=g''(\alpha)$.

Daniel R
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hamam_Abdallah
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  • Thank you for posting an Answer, can you confirm that your $e_{n+1}$ takes into account my recent edit: $x_{n+1} = x_n - \frac{f(x_n)}{f(x_0)}$? – Ozera Sep 29 '16 at 12:09
  • @Salahamam In the answers of my guide book it comes out that $C=1-f''(\xi_n)/f'(x_0), s=1$, could you explain to me why? – user482152 Mar 08 '18 at 02:12