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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?

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  • $\begingroup$ Do they give you a starting point for the iteration? $\endgroup$ – Amzoti Aug 4 '13 at 2:06
  • $\begingroup$ @Amzoti No, I have not been given a starting point. $\endgroup$ – ABC Aug 4 '13 at 2:08
  • $\begingroup$ See my solution and just pick one as it does not matter where you start, you should find the 5th root of 3 to three places. $\endgroup$ – Amzoti Aug 4 '13 at 2:08
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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 $3^{1/5} = 1.2457309396155173259...$

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  • $\begingroup$ That is, i am wrong for the first part of the question.I made mistake thinking $a$ as the root . It will be $a^{\frac{1}{3}}=x$ where $x$ is the root. Isn't it? $\endgroup$ – ABC Aug 4 '13 at 2:18
  • $\begingroup$ @harry, it doesn't matter what we call the roots $a$, $x$, either works. I like to keep things straight with $f(x)$, but call it what you will. That does not change the fact that we need the fifth root. Is this clear? Regards $\endgroup$ – Amzoti Aug 4 '13 at 2:19
  • $\begingroup$ Greetings, dear friend! This might be dinner time for you ;-) $\endgroup$ – amWhy Aug 6 '13 at 0:23

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