Single Variable Calculus, Chapter 4, 4.8, Section 4.8, Problem 12

Use Newton's Method to approximate $\sqrt[100]{100}$ correct to eight decimal places. $\sqrt[100]{100}$ is equal to the positive root of $x^{100} - 100 = 0$

So we take $f(x) = x^{100} - 100$. Then


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{d}{dx} (x^{100}) - \frac{d}{dx} (100)
\\
\\
f'(x) =& 100x^{99}

\end{aligned}
\end{equation}
$


Using Newton's Method

$\displaystyle x_{n + 1} = x_n - \frac{x_n^{100} - 100}{100x_n^{99}}$

If we choose $x_1 = 1.05$ as initial approximation, then we have


$
\begin{equation}
\begin{aligned}

x_2 \approx & 1.04748471
\\
\\
x_2 \approx & 1.04713448
\\
\\
x_4 \approx & 1.04712855
\\
\\
x_5 \approx & 1.04712855

\end{aligned}
\end{equation}
$



Since $x_4$ and $x_5$ agree to eight decimal places, therefore $\sqrt[100]{100} \approx 1.04712855$ to eight decimal places.