how to solve The n-th root of a raised to the power of m in newton way and then solve it for The third root of 5 raised to the power of 1 with with a tolerance of 10^-4  and x0=1.5

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).

---

### 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.

how to solve The n-th root of a raised to the power of m in newton way and then solve it for The third root of 5 raised to the power of 1 with with a tolerance of 10^-4 and x0=1.5

Learn how to solve the third root of 5 using Newton's Method. This step-by-step guide covers numerical methods for root finding, including detailed iterations with a tolerance of 10^-4 and an initial guess of 1.5. Ideal for college-level students studying numerical methods or calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation