Question

Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.) x5 - x - 7 = 0, x1 = 1

Answer 1

To find x3, the third approximation to the root of the equation x^5 - x - 7 = 0 using Newton's method and the initial approximation x1 = 1, we follow these steps: calculate f(x1), calculate f'(x1), use the formula x2 = x1 - f(x1)/f'(x1) to find the second approximation, and repeat steps 1-3 to find x3. The third approximation to the root is x3 = 1.7501 (rounded to four decimal places).

Step-by-step explanation:

Newton's method is an iterative method for finding the roots of a given equation. It works by using an initial approximation and repeatedly refining it to get better and better approximations to the root. In this case, we are given the equation x^5 - x - 7 = 0 and the initial approximation x1 = 1. To find x3, the third approximation to the root, we follow these steps:

1. Calculate f(x1), which is the value of the equation at x1. In this case, f(x1) = x1^5 - x1 - 7 = 1^5 - 1 - 7 = -7.
2. Calculate f'(x1), which is the derivative of the equation evaluated at x1. In this case, f'(x1) = 5x1^4 - 1 = 5(1)^4 - 1 = 4.
3. Use the formula x2 = x1 - f(x1)/f'(x1) to find the second approximation. Substituting the values we have, x2 = 1 - (-7)/4 = 1 + 7/4 = 1.75.
4. Repeat steps 1-3 to find x3, the third approximation. Using x2 as the new initial approximation, we have f(x2) = x2^5 - x2 - 7 = 1.75^5 - 1.75 - 7 = -0.03125, f'(x2) = 5x2^4 - 1 = 5(1.75)^4 - 1 = 23.515625. Substituting these values into the formula, x3 = 1.75 - (-0.03125)/23.5156 = 1.7501.

Therefore, the third approximation to the root of the equation x^5 - x - 7 = 0, using Newton's method and the initial approximation x1 = 1, is x3 = 1.7501 (rounded to four decimal places).