Use Newton\'s method to find the second and third approximation of a root of x^3

Question

Use Newton's method to find the second and third approximation of a root of x^3+x+2=0 starting with x prime 1= -1 as the initial approximation. What are the second and third approximations?

Explanation / Answer

cos(x^2 - 7) = x^3 ==> x^3 - cos(x^2 - 7) = 0. This implies f(x) = x^3 - cos(x^2 - 7) and: f'(x) = 3x^2 + 2x*sin(x^2 - 7). By Newton's Method, the following recursive formula will give us a better approximation for the root of the function: x_n+1 = x_n - f(x_n)/f'(x_n). In this case: x_n+1 = x_n - [(x_n)^3 - cos(x_n^2 - 7)]/[3(x_n)^2 + 2(x_n)*sin(x_n^2 - 7)]. With x1 = 1, we can obtain x2 ˜ 0.988808 and x3 ˜ 0.988620.

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