To solve for the $n$-th root of $a^m$ using Newton's Method, we need to break down the problem into steps. In general, Newton's Method is an iterative approach to solving equations of the form $f(x) = 0$. We start with an initial guess and improve it iteratively by using the formula:

$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$

1. Problem Setup for $\sqrt[n]{a^m}$

The $n$-th root of $a^m$ can be written as:

$x = (a^m)^{1/n} = a^{m/n}$

To find this value using Newton's method, we want to solve for $x$ in the equation:

$x^n - a^m = 0$

This is the equation we want to solve using Newton's method. So we define the function:

$f(x) = x^n - a^m$

and its derivative:

$f'(x) = n x^{n-1}$

2. Newton's Iterative Formula

Using the above expressions for $f(x)$ and $f'(x)$, Newton's iterative formula becomes:

$x_{k+1} = x_k - \frac{x_k^n - a^m}{n x_k^{n-1}}$

3. Example: Solve $\sqrt[3]{5^1}$ with tolerance $10^{-4}$ and $x_0 = 1.5$

Here we want to solve for the third root of 5 (i.e., $n = 3$, $a = 5$, and $m = 1$).

• The equation we are solving is: $x^3 - 5 = 0$.
• The iterative Newton formula becomes:

$x_{k+1} = x_k - \frac{x_k^3 - 5}{3 x_k^2}$

We will perform the iterations until the change between successive values of $x$ is smaller than the tolerance of $10^{-4}$.

4. Iterations

Start with $x_0 = 1.5$ and apply the formula:

1. Iteration 1: $x_1 = 1.5 - \frac{1.5^3 - 5}{3 \times 1.5^2} = 1.5 - \frac{3.375 - 5}{6.75} = 1.5 + 0.2407 = 1.7407$

2. Iteration 2: $x_2 = 1.7407 - \frac{1.7407^3 - 5}{3 \times 1.7407^2} = 1.7407 - \frac{5.278 - 5}{9.0986} = 1.7407 - 0.0306 = 1.7101$

3. Iteration 3: $x_3 = 1.7101 - \frac{1.7101^3 - 5}{3 \times 1.7101^2} = 1.7101 - \frac{4.99856 - 5}{8.7727} = 1.7101 + 0.000164 = 1.7103$

4. Iteration 4: $x_4 = 1.7103 - \frac{1.7103^3 - 5}{3 \times 1.7103^2} = 1.7103 - \frac{5.00001 - 5}{8.774} = 1.7103 - 0.000001 = 1.7103$

At this point, the change between successive iterations is less than $10^{-4}$, so we stop.

Final Solution:

The third root of 5 is approximately $x = 1.7103$ (to four decimal places).

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Any Questions?

Would you like more details on Newton's method, or do you have any specific questions?

Related Questions:

1. How can Newton's method be used to solve general polynomial equations?
2. How does the choice of the initial guess $x_0$ affect the convergence of Newton's method?
3. What is the difference between Newton's method and other root-finding algorithms?
4. Can Newton's method fail? Under what conditions does it not converge?
5. How can Newton's method be adapted for complex roots?

Tip:

Always check the derivative at each iteration in Newton's method. If the derivative becomes too small, the method might fail to converge.