An exercise of Newton-Raphson Method .

Question

Derive the Newton-Raphson iteration formula for $a^{\frac{1}{5}}$ where $a$ is a real positive number and then find $3^{\frac{1}{5}}$ correct to $3$ decimal places.

My attempt:

$f(a)=a^{\frac{1}{5}}$

$f'(a)=\frac{1}{5}a^{-\frac{4}{5}}$

The Newton-Raphson iteration formula:

$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}=a_n-\frac{a^{\frac{1}{5}}}{\frac{1}{5}a^{-\frac{4}{5}}}=-4a_n$$

So each guess is $-4$ times the previous guess. Then clearly the guesses diverge and the Newton-Raphson method fails.

Then how can I find $3^{\frac{1}{5}}$ correct to $3$ decimal places? What will I conclude here? Am I correct for the first half of the question?

Answer 1

Hint:

What you need is a formula to get the $5^{th}$ root of $3$.

So, we have:

$$x^5 = a \rightarrow f(x) = x^5 - a$$

Can you repeat the process for the Newton solution again with $a = 3$?

Hover over this area to see more details.

The Newton iteration is given by: $$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)} = x_n - \dfrac{x^5_n - 3}{5x^4_n}$$ $x_0 = 1$, $x_1 = 1.4$, $x_2 = 1.27618492295$, $x_3 = 1.24715013208$, $x_4 = 1.24573416588$, $x_5 = 1.24573093963$, Thus, the answer to 3-places is $3^{1/5} = 1.246$. You can easily check this solution as $3^{1/5} = 1.2457309396155173259...$