Given a sequence for the cubic root of a number $Y=\sqrt[3]{X}$

Question

Given a sequence for the cubic root of a number $Y=\sqrt[3]{X}$, if $a>0$, show that Y always lies between a and $X/a^2$ (if $a<Y$, then $X/a^2 > Y$, etc)

I'm thinking use Newton's Method for approximation, use $a_{n+1}=a_n- \frac{f(a_n)}{f'(a_n)}$, I end up with $a_{n+1}=a_n-3a_n=-2a_n$. And I can't proceed, can someone tell me if I'm going the wrong direction?

Answer 1

Newton's Method does not work in this case because there are certain hypotheses on the derivative that have to be satisfied in order for that method to work.

This looks like an application of the intermediate value theorem. Have you tried that? Or are you looking for another method?

Update: Let $f(x) = x^3 - X$. If $a < Y$ then $f(a) = a^3 - Y^3 < 0$, and $f(X/a^2) = X^3 / a^6 - X > X^3 / Y^6 - X = X^3 / X^2 - X = 0$.

Similar argument if $a > Y$.

And don't forget that $f$ is continuous because it's a polynomial.

Answer 2

To use Newton's method to approximate $X^{1/3}$ for $X\ne 0$ let $g(y)=y^3-X.$ Let $y_0 \ne 0$ and let $y_{n+1}=y_n-g(y_n)/g'(y_n)=(2 y_n^3+X)/3 y_n^2.$ Then $y_n$ converges to $X^{1/3}.$ A modification of this is $Y_{n+1}=Y_n- Y_ng(Y_n)/(2 Y_n^3+X)$ which converges faster.