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Newton's method for $\frac1{\sqrt{a}}$ proceeds by iterating minimizing $\frac1{x^2}-a$.

Why cant I do $\frac{x-ax^3}2$?

Infact why cant I replace $2$ by any $k\in\Bbb R_{>0}$?

Turbo
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  • Why do you need to divide by 2? Does that change the minima (if there are any)? – avs Dec 02 '16 at 21:25
  • Because you introduce another zero to the problem $x = 0$, which complicates things numerically – caverac Dec 02 '16 at 21:44
  • Please reformulate your question. Your term as function does not have a minimum, the root of your term is $\frac1{\sqrt{a}}$ which indeed Newton's method will find. – Lutz Lehmann Dec 02 '16 at 22:14
  • @LutzL $\frac{\frac1{\sqrt a}-a\frac1{\sqrt{a^3}}}k=\frac{\frac1{\sqrt a}-\frac1{\sqrt a}}k=0$. – Turbo Dec 02 '16 at 22:19
  • That's what I said, your term is zero at $\frac1{\sqrt{a}}$. Which contradicts at least two statements in your question. – Lutz Lehmann Dec 02 '16 at 22:24
  • @LutzL ok corrected the mistake. – Turbo Dec 02 '16 at 22:32

1 Answers1

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Using $f(x)=x^d-ax^{2+d}$ the Newton method iterates $$ x_{\it next} = x - \frac{x^d-ax^{2+d}}{dx^{d-1}-(d+2)ax^{d+1}} =x·\frac{(d+1)ax^2-(d-1)}{(d+2)ax^2-d} $$ where for $d=0$, $d=-2$ and $d=-\frac12$ you will rediscover some well-known methods for square root approximation.

Lutz Lehmann
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  • So both are right? – Turbo Dec 02 '16 at 22:22
  • I dont follow your answer. – Turbo Dec 02 '16 at 22:24
  • That depends on what the question was. The methods differ in their basin of convergence, restriction on the use of division and order of convergence. – Lutz Lehmann Dec 02 '16 at 22:26
  • could you please specifically comment on my two expressions? – Turbo Dec 02 '16 at 22:31
  • There are no equations in your question. The can be easily identified by the equality sign $=$ which is not to be found in the entire text. – Lutz Lehmann Dec 02 '16 at 22:36
  • Please explain in more words or formulas what you mean with "replace $2$ by any $k$". And how someone or something is trying to prevent you from doing it. It is perfectly legal to consider instead of $f(x)=0$ the equation $f(x)·g(x)=0$. One needs to be aware that the roots and poles of $g$ are added to the situation. The Newton step for the modified function is $$x_{\rm next}=x-\frac{f(x)g(x)}{f'(x)g(x)+f(x)g'(x)}=x-\frac{f(x)}{f'(x)+f(x)·(\ln|g(x)|)'},$$ which is also perfectly fine, but will give different iteration sequences than for the unmodified function. – Lutz Lehmann Dec 03 '16 at 01:14