Use Newton's Method to approximate 12 to 6 significant figures starting with 20 = 3.5. Compare with the value obtained f
Posted: Tue May 10, 2022 9:16 pm
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Sometimes Newton's method does not work. This exercise illustrates an example, Use Newton's method to find a root of p(x) = 23 – , with the starting value xo = 15 and observe what happens with the first few iterations. For your calculations, use exact arithmetic (not decimals). For example, enter so as sqrt(5)/5. 20 = 21 22= 23 = 24 Newton's method may fail in many different ways. A detailed analysis of Newton's method, and related methods, is a huge subject and well beyond the scope of our course.
The number 16 can be thought of as the solution of the equation x5 – 6 = 0. Newton's method is an iterative numerical method to find roots (zeroes) of an equation of of the form f(x) = 0. To implement Newton's method, one starts with an initial approximation to and then computes a sequence of f(3) approximations via the formula tk+1 = g(Tk), where g(x) = 3 - f'(x) To find V6, the iteration function is: g(x) = With the starting value xo = 1, the first few iterates are: 21 = 32 = 23 = 24 = 25 =
Use a linear approximation (i.e., linearization) to approximate sin (44%): Let f(x) = sin(x). Find the equation of the line tangent to f(x) at a "nice" point near 44°. Then, use the tangent line to approximate sin (44). sin(44) Hint: Convert degrees to radians in your calculations.