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Consider the solution of the following nonlinear equation using Newton's method. 2.7 x3 = 5cos() +13 After two iterations, using ten digit floating point arithmetic,
Consider the solution of the following nonlinear equation using Newton's method. 2.7 x3 = 5cos() +13 After two iterations, using ten digit floating point arithmetic, the estimated solution is: X2 =1.7250997086 Using Newton's method find the next estimate of the solution and use this to determine an absolute estimate of the error in X2. Maintain at least TEN significant digits throughout all your calculations. When entering your results you may round your answers to eight decimal digits. For example x3 = 2.12345678, |Error in X2| $ 0.00001234 X3 = |Error in X2| S
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