So I've been preparing for an exam by looking over old exam papers and I came across this question:
a) Using Newton's method, show that the iteration equation to find the cube root of $R$ can be written as
$$x_{n+1}= \frac 13\left (2x_n+ \frac R{x_n^2}\right)$$
b) If $R=3$ and $x_0=1$, calculate $x_1$ and $x_2$.
c) The equation in (a) is used to approximate the cube root of $R$. Using the values in (b), calculate the exact relative error in the approximation of $x_2$.
I didn't have any troubles with showing the iteration equation from the Newton's method formula for part (a), and here's what I did for part (b):
$x_1=\frac 13\left (2x_0 +\frac R{x_0^2}\right)$
$x_1=\frac 13\left (2.1 +\frac 3{1^2}\right) = \frac 53$
$x_2=\frac 13\left (2x_1 +\frac R{x_1^2}\right)$
$x_2=\frac 13\left (2\frac 53 +\frac 3{\frac 53^2}\right) = \frac {77}{45}$
which I'm confident is right (but please correct me if I'm wrong). And here's my attempt at part (c):
$x=\sqrt[3] 3$, $\hat x=x_2=\frac {77}{45}$
$e_R=\left\lvert \frac {x-\hat x}{x} \right\rvert$
$e_R=\left\lvert \frac {\sqrt[3] 3-\frac {77}{45}}{\sqrt[3] 3} \right\rvert = 0.1864181806$
I was just hoping if someone would be able to tell me if this answer is the correct answer for what part (c) is asking. In my subject notes there's no mention of "exact relative error", just relative error, and I was thinking that the exact part just referred to the fact that we know what the cube root of 3 equals. Googling "exact relative error" didn't help, so if someone is just able to tell me if they think what I've done is right and perhaps shed some light on "exact relative error" that'd be great. Thanks in advance.
Best Answer
No it is not correct, because your $x_2$ is in error. $x_1$ is correct but $$x_2=\frac 13\left (2x_1 +\frac R{x_1^2}\right)=\frac 13\left (2\frac{5}{3} +\frac {3}{(\frac{5}{3})^2}\right)=\frac{1}{3}\left (\frac{10}{3} +\frac{27}{25}\right) =\frac{331}{225}\cdot$$
And therefore $$e_R=\left\lvert \frac {\sqrt[3] 3-\frac {331}{225}}{\sqrt[3] 3} \right\rvert \approx 0.0200114747$$