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approximated root up to six decimal place is -1.249052. In x 2 and x 3 6 decimal places are same. then successive iteration from above equations are Let x 0 = −1.1 be the initial approximation. Now, since f (−1) 0, root lies in between −1 and −2. Then Newton-Raphson formula for this problem reduces to Solution: Let f (x) = sin x − 1 − x 3, then f ′(x) = cos x − 3x 2. This result is known as Newton-Raphson formula.Įxample: Solve sinx = 1 x 3 using Newton-Raphson Method. We may iterate the process to refine the root. Since h is small, we can neglect second, third and higher degree terms in h and thus we get Let x 0 be an approximation of a root of the given equation f (x) = 0, Let x 1 = x 0 h be the exact approximation of the root. When an approximate value of a root of an equation is given, this method is used to obtain better and closer approximation to the root.

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