I have a text which claims the following (a) is Newton's method for cube roots, where $y$ is an approximation to the cube root of $x$:
It's my understanding the form can be derived from:
where each $x_n$ is a better approximation of the root with each iteration and $x_0$ being an initial approximate guess.
I'm confused when deriving form (a). Can anyone explain how that form is obtained? My derivation shows:
$f(x) = x^3 - a$ and $f'(x) = 3x^2$ where $a$ is the cube, then we find $x$ for $f(x) = 0$
$x_1 = x_0 - ((x_0)^3 - a)/(3(x_0)^2)$
UPDATE: One of the derivatives was calculated incorrectly. When the numerator and denominator are divided now by $x^2$ instead, the problem is resolved.
Thank you
$\endgroup$ 13 Answers
$\begingroup$In the (k+1)-th iteration$$x_{k+1} = x_k- \frac{x_k^3-a}{3x_k^2}=x_k - \frac{1}{3}x_k + \frac{a}{3x_k^2} = \frac{a/x_k^2 + 2x_k}{3}$$
$\endgroup$ $\begingroup$One point on the curve $y=x^3-a$ is $P_n=(x_n,x_n^3-a)$. By definition, $x_{n+1}$ is the X-coordinate (or "abscissa" if you wish to impress your friends, relatives and neighbours) of the point of intersection of the tangent line to the curve at the point $P_n$ with the X-axis. The slope of the tangent line is the derivative $y'=3x_n^2$ when $x=x_n.$ Thus the equation of the tangent line is $$\frac{y-(x_n^3-a)}{x-x_n}=3x_n^2$$ So take this equation, set $y=0$ and solve for $x.$ By definition, this value of $x$ is $x_{n+1}$ So $$\frac{-(x_n^3-a)}{x_{n+1}-x_n}=3x_n^2$$ and thus $$x_{n+1}=x_n-\frac{x_n^3-a}{3x_n^2}$$
$\endgroup$ $\begingroup$UPDATE: One of the derivatives was calculated incorrectly. When the numerator and denominator are divided now by $x^2$ instead, the problem is resolved.
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