2
$\begingroup$

Show how you could use the Newton-Raphson method on $f(x) = x^3 - a$ to calculate $a^{1/3}$. Does it converge? Prove your answer.

This is what I have:

If $x^*$ is a root of $f(x)$, then $f(x^*) = 0 \Rightarrow (x^*)^3 -a = 0 \Rightarrow x^* = a^{1/3}$. Thus $x^* = a^{1/3}$ is the only root of the function $f(x)$.

To calculate the root $a^{1/3}$ we can use Newton-Raphson,

$\displaystyle{x_{n+1} = x_{n} -\frac{(x_n)^3 -a}{3(x_n)^2}}$

How can I prove that this converges or fails to converge?

$\endgroup$
  • $\begingroup$ For $a>0:$ Suppose $a^{1/3}\ne x_1>0 .$ Then $x_2>a^{1/3}.$ And if $x_n>a^{1/3}$ then $x_n>x_{n+1}>a^{1/3}.$ So $(a_n)_n$ is a descending sequence bounded below by $a^{1/3}$ so it has a limit $L\ne 0. $ Then $L=\lim_{n\to \infty} a_{n+1}=$ $\lim_{n\to \infty}(a_n-(a_n^3-a)/3a_n^2)=$ $L-(L^3-a)/3L^2,$ which implies $L^3-a=0.$... Graphically the tangent-line to the curve $y=x^3-a$ at the point $(x_n, x_n^3-a) $ intersects the $x$-axis at $(x_{n+1},0).$ $\endgroup$ – DanielWainfleet Nov 1 '17 at 3:21
  • $\begingroup$ More generally, for $a>0$ and $x_2>a^{1/3}.$ The function $f(x)=x^3-a$ is positive for $a^{1/3}<x\leq x_2$, and $f'(x)>0$ for $a^{1/3}\leq x\leq x_2.$ By the "graphical " property in my previous comment, this implies $a^{1/3}<x_{n+1}<x_n$ for $n\geq 2.$ The limit $L$ of $(x_n)_n$ satisies $L=L-f(L)/f'(L)$, so $f(L)=0$. $\endgroup$ – DanielWainfleet Nov 1 '17 at 3:30
  • $\begingroup$ ERRATUM, In my previous comment the property $(x\in [a^{1/3},x_2]\implies f'(x)>0)$ should be replace by ($f'$ is strictly increasing for $x\in [a^{1/3},x_2])$ in order to use the MVT (and the positivity of $f(x)$ on $(a^{1/3}.x_2])$ to show that $a^{1/3}<x_{n+1}<x_n$ for $n\geq 2$ (regardless of what $f$ actually is). $\endgroup$ – DanielWainfleet Nov 1 '17 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.