Math Problem Statement

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.

Solution

To derive a formula using the Newton-Raphson method for finding the kkth root of a positive number NN, 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: xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

For finding the kkth root of NN, we want to find xx such that xk=Nx^k = N. We can reformulate this as finding the root of the function: f(x)=xkNf(x) = x^k - N

The derivative of f(x)f(x) is: f(x)=kxk1f'(x) = kx^{k-1}

Using the Newton-Raphson formula, we have: xn+1=xnxnkNkxnk1x_{n+1} = x_n - \frac{x_n^k - N}{kx_n^{k-1}}

Simplifying the right-hand side: xn+1=xnxnkNkxnk1=xnxnkkxnk1+Nkxnk1=xnxnk+Nkxnk1x_{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: xn+1=(k1)xn+Nxnk1kx_{n+1} = \frac{(k-1)x_n + \frac{N}{x_n^{k-1}}}{k}

Thus, the iterative formula to find the kkth root of NN using the Newton-Raphson method is: xn+1=(k1)xn+Nxnk1kx_{n+1} = \frac{(k-1)x_n + \frac{N}{x_n^{k-1}}}{k}

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

Here, N=25N = 25 and k=4k = 4.

  1. Choose an initial guess, x0x_0. Let's choose x0=2x_0 = 2.

  2. Apply the iterative formula: xn+1=(41)xn+25xn414=3xn+25xn34x_{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: x1=32+25234=6+2584=6+3.1254=9.1254=2.28125x_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: x2=32.28125+252.281253432.28125+2511.90246.84375+2.14=8.9437542.2359375x_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: x3=32.2359375+252.23593753432.2359375+2511.17046.7078125+2.2394=8.946812542.236703125x_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)1/42.236(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.

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Math Problem Analysis

Mathematical Concepts

Newton-Raphson method
Root finding

Formulas

Newton-Raphson iterative formula for root finding

Theorems

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Suitable Grade Level

Advanced High School to College