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While on the Internet I came across a formula for cube root using recursion. The formula was:

$$ use \space x_1 = \frac{a}{3} \space (a \space is \space the \space number \space we \space want \space to \space find \space cube \space root \space of)$$

Then use the formula:

$$ x_{n+1} = \frac 13 \left( 2x_n + \frac{a}{{x_n}^2} \right)$$

Recursively.

What is this method and how does it work. Is their someway I can prove it ?

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  • $\begingroup$ The formula is wrong. You need to divide the RHS by $3$. $\endgroup$ – Yves Daoust Feb 16 '17 at 14:34
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If you want to solve for $x$, consider $$f(x)=x^3-a\implies f'(x)=3x^2$$ Now, using Newton method $$x_{n+1}=x_ n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^3-a}{3x_n^2}=\frac{2x_n^3+a}{3x_n^2}=\frac 23x_ n+\frac a {3x_n^2}$$

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  • $\begingroup$ There doesn't appear to be a factor of $2$ that can be removed from $a$ in the last equality. $\endgroup$ – Michael Burr Feb 16 '17 at 14:46
  • $\begingroup$ @MichaelBurr. Superb typo !!! Thanks for pointing it. $\endgroup$ – Claude Leibovici Feb 16 '17 at 15:23
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Suppose that $a \ne 0.$ We have that $$ x^3=a \iff x =\frac 13 \left( 2x+\frac a{x^2} \right). $$ Thus the problem is reduced to finding the fixed point of the function $$ f(x)=\frac 13 \left( 2x+\frac a{x^2} \right), $$ which can be done with the use of the method of fixed-point iterations. Google the phrase "method of fixed-point iterations".

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