Question

5. Newton's Method a. Discuss the use of Newton's method to approximate solutions to a system of n nonlinear equations with n unknowns. b. Write the linear system of equations given by -200u+ 100u, = sin(0.1) 1001 - 200u2 + 100u3 = sin(0.2) 1002 - 200uz + 100u4 = sin(0.3) 100u3 - 2004 + 100us = sin(0.4) 100u4 - 200us + 100u = sin(0.5) 100us - 2004g + 100u, = sin(0.6) 1006 - 2004; + 100ug = sin(0.7) 100u; - 200ug + 100u, = sin(0.8) 100ug - 2004, = sin(0.9) in the form Au = b where A E R9X9 and b E R are known, and u= ( u u )" is unknown. c. Solve the system in part b. d. Use the solution found in part c as an initial guess with Newton's method and a = 0.1, to obtain a good estimate for the solution to the nonlinear system -200u+ 100u2 = sin(0.1) + au ? 100u - 200uz + 100uz = sin(0.2) + auz? 100u2 - 200uz + 100u4 = sin(0.3) + auz? 100u3 - 2004 + 100u = sin(0.4) + aug? 1004 - 200us + 100u = sin(0.5) + aus? 100us - 2000g + 100u, = sin(0.6) + aus? 100u - 2004, + 100ug = sin(0.7) + aus? 100u, - 2004g + 100u, = sin(0.8) + aug? 100ug - 200u, = sin(0.9) + au, Note: You might have to do several iterations of Newton's method. e. Use the approximation obtained in part d as an initial guess for Newton's method in the case a = 0.2, and continue this process until you have obtained good approximations for solutions to this system of equations for a = i/10, for i = 0,...,10.

pls answer e.

Answer 1

Newton's Method is a mathematical tool often used in numerical analysis, which serves to approximate the zeroes or roots of a function (that is, all x:f(x)=0).

The method is constructed as follows: given a function f(x) defined over the domain of real numbers x, and the derivative of said function (f'(x)), one begins with an estimate or "guess" as to where the function's root might lie. For example, suppose one is presented with the function f(x)=x2+x−2.5. This is similar to another function g(x)=x2+x−2, whose roots are x=1 and x=−2. Thus, thanks to this similarity, one might use x=1 or x=−2 as guesses to start Newton's Method with f(x).

(Alternately, if a graphical representation is available but the exact root is not listed, an acceptable approximation might be the nearest whole number to the root).

Whatever method used, we declare this initial guess to be x0. We arrive at a better approximation, x1, by employing the Method: x1=x0−f(x0)f'(x0). Essentially, by utilizing the derivative, one is able to increment closer to the actual value. In the above example, f(x)=x2+x−2.5, if we assume x0=1, then x1=1−f(1)f'(1)=1−−.53=76or≈1.16667.

Often, one may be able to find the root another way (by using a graphing calculator, for example), and an exam item or textbook problem may demand a certain degree of accuracy (such as within 1% of the actual value). In such a case, if x1 is not an accurate enough approximation, one performs the iteration again, as often as needed for the desired degree of accuracy. The formula to find the general xn, then, is xn=xn−1−f(xn−1)f'(xn−1)