Using Newton−Raphson method, derive a formula
for finding the kth root of a positive number N and
hence compute the value of (25)1/4.

To derive a formula using the Newton-Raphson method for finding the \(k\)th root of a positive number \(N\), let's start with the general approach.

The Newton-Raphson method is used to find successively better approximations to the roots (or zeroes) of a real-valued function. The method is based on the formula:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

For finding the \(k\)th root of \(N\), we want to find \(x\) such that \(x^k = N\). We can reformulate this as finding the root of the function:
\[ f(x) = x^k - N \]

The derivative of \(f(x)\) is:
\[ f'(x) = kx^{k-1} \]

Using the Newton-Raphson formula, we have:
\[ x_{n+1} = x_n - \frac{x_n^k - N}{kx_n^{k-1}} \]

Simplifying the right-hand side:
\[ x_{n+1} = x_n - \frac{x_n^k - N}{kx_n^{k-1}} = x_n - \frac{x_n^k}{kx_n^{k-1}} + \frac{N}{kx_n^{k-1}} = x_n - \frac{x_n}{k} + \frac{N}{kx_n^{k-1}} \]

Combining the terms gives us:
\[ x_{n+1} = \frac{(k-1)x_n + \frac{N}{x_n^{k-1}}}{k} \]

Thus, the iterative formula to find the \(k\)th root of \(N\) using the Newton-Raphson method is:
\[ x_{n+1} = \frac{(k-1)x_n + \frac{N}{x_n^{k-1}}}{k} \]

### Example: Compute \((25)^{1/4}\)

Here, \(N = 25\) and \(k = 4\).

1. Choose an initial guess, \(x_0\). Let's choose \(x_0 = 2\).

2. Apply the iterative formula:
\[ x_{n+1} = \frac{(4-1)x_n + \frac{25}{x_n^{4-1}}}{4} = \frac{3x_n + \frac{25}{x_n^3}}{4} \]

Let's perform a few iterations:

- **Iteration 1:**
  \[ x_1 = \frac{3 \cdot 2 + \frac{25}{2^3}}{4} = \frac{6 + \frac{25}{8}}{4} = \frac{6 + 3.125}{4} = \frac{9.125}{4} = 2.28125 \]

- **Iteration 2:**
  \[ x_2 = \frac{3 \cdot 2.28125 + \frac{25}{2.28125^3}}{4} \approx \frac{3 \cdot 2.28125 + \frac{25}{11.902}}{4} \approx \frac{6.84375 + 2.1}{4} = \frac{8.94375}{4} \approx 2.2359375 \]

- **Iteration 3:**
  \[ x_3 = \frac{3 \cdot 2.2359375 + \frac{25}{2.2359375^3}}{4} \approx \frac{3 \cdot 2.2359375 + \frac{25}{11.170}}{4} \approx \frac{6.7078125 + 2.239}{4} = \frac{8.9468125}{4} \approx 2.236703125 \]

Further iterations will yield results closer to the true value. After a sufficient number of iterations, we will approximate \((25)^{1/4} \approx 2.236\).

Would you like more details or have any questions?

Here are 5 questions you might want to explore next:
1. How can the convergence of the Newton-Raphson method be proven?
2. Can the Newton-Raphson method be applied to functions with multiple variables?
3. How does the choice of initial guess affect the convergence of the Newton-Raphson method?
4. What are some practical applications of the Newton-Raphson method in engineering?
5. Can the Newton-Raphson method be used for functions that are not differentiable?

**Tip:** When choosing the initial guess for the Newton-Raphson method, select a value close to the expected root to ensure faster convergence.

Learn how to compute the 4th root of 25 using the Newton-Raphson method with detailed step-by-step calculations. This problem involves applying the iterative formula for root finding, suitable for advanced high school to college-level students.

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