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In this assignment we will introduce a standard method to find approximate solutions to nonlinear equations. Suppose that f is a nonlinear differentiable function and that we wish to solve the equation f(x) = 0. Recall that solving f(x) = 0 is equivalent to finding an x-intercept of this graph and that such an equation may have no solutions or many solutions. The method we outline here, called Newton's Method, is an approach for finding just one solution (if such a solution exists). Newton's Method starts with an initial guess for the solution, xo, and improves upon it as follows: 1. Calculate the tangent line to y = f(x) at (xo, f(x0)) 2. Find the x-intercept of the tangent line and set this equal to 11. This process is illustrated graphically in figure 1. If the conditions are right, a₁ will be a better approximation to the solution than the previous one, xo, and repeating this process with the new approximation ought to improve it again. Iterating this process will give a sequence of approximations which should approach the actual solution. Tangent at (xo, f(x0)) Actual solution (unknown) (xo, f(xo)) 21 20 y = f(x) Figure 1: The first step of Newton's Method. The next step would involve calculating a tangent line at (x1, f(x1)) and finding the x-intercept 2. 1 1. Recall that the equation of a line with gradient m passing through the point (a, b) is y ẞ m(x-a). Show that if xn+1 is the x-intercept of the tangent to f at (xn, f(xn)), then Xn+1 Xn - f(xn) f'(xn) provided that f'(xn) ± 0. (1) [3 marks]

In this assignment we will introduce a standard method to find approximate solutions to nonlinear equations. Suppose that f is a nonlinear differentiable function and that we wish to solve the equation f(x) = 0. Recall that solving f(x) = 0 is equivalent to finding an x-intercept of this graph and that such an equation may have no solutions or many solutions. The method we outline here, called Newton's Method, is an approach for finding just one solution (if such a solution exists). Newton's Method starts with an initial guess for the solution, xo, and improves upon it as follows: 1. Calculate the tangent line to y = f(x) at (xo, f(x0)) 2. Find the x-intercept of the tangent line and set this equal to 11. This process is illustrated graphically in figure 1. If the conditions are right, a₁ will be a better approximation to the solution than the previous one, xo, and repeating this process with the new approximation ought to improve it again. Iterating this process will give a sequence of approximations which should approach the actual solution. Tangent at (xo, f(x0)) Actual solution (unknown) (xo, f(xo)) 21 20 y = f(x) Figure 1: The first step of Newton's Method. The next step would involve calculating a tangent line at (x1, f(x1)) and finding the x-intercept 2. 1 1. Recall that the equation of a line with gradient m passing through the point (a, b) is y ẞ m(x-a). Show that if xn+1 is the x-intercept of the tangent to f at (xn, f(xn)), then Xn+1 Xn - f(xn) f'(xn) provided that f'(xn) ± 0. (1) [3 marks]

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In this assignment we will introduce a standard method to find approximate solutions to nonlinear equations. Suppose that f is a nonlinear differentiable function and that we wish to solve the equation f(x) = 0. Recall that solving f(x) = 0 is equivalent to finding an x-intercept of this graph and that such an equation may have no solutions or many solutions. The method we outline here, called Newton's Method, is an approach for finding just one solution (if such a solution exists). Newton's Method starts with an initial guess for the solution, xo, and improves upon it as follows: 1. Calculate the tangent line to y = f(x) at (xo, f(x0)) 2. Find the x-intercept of the tangent line and set this equal to 11. This process is illustrated graphically in figure 1. If the conditions are right, a₁ will be a better approximation to the solution than the previous one, xo, and repeating this process with the new approximation ought to improve it again. Iterating this process will give a sequence of approximations which should approach the actual solution. Tangent at (xo, f(x0)) Actual solution (unknown) (xo, f(xo)) 21 20 y = f(x) Figure 1: The first step of Newton's Method. The next step would involve calculating a tangent line at (x1, f(x1)) and finding the x-intercept 2. 1
Transcribed Image Text:In this assignment we will introduce a standard method to find approximate solutions to nonlinear equations. Suppose that f is a nonlinear differentiable function and that we wish to solve the equation f(x) = 0. Recall that solving f(x) = 0 is equivalent to finding an x-intercept of this graph and that such an equation may have no solutions or many solutions. The method we outline here, called Newton's Method, is an approach for finding just one solution (if such a solution exists). Newton's Method starts with an initial guess for the solution, xo, and improves upon it as follows: 1. Calculate the tangent line to y = f(x) at (xo, f(x0)) 2. Find the x-intercept of the tangent line and set this equal to 11. This process is illustrated graphically in figure 1. If the conditions are right, a₁ will be a better approximation to the solution than the previous one, xo, and repeating this process with the new approximation ought to improve it again. Iterating this process will give a sequence of approximations which should approach the actual solution. Tangent at (xo, f(x0)) Actual solution (unknown) (xo, f(xo)) 21 20 y = f(x) Figure 1: The first step of Newton's Method. The next step would involve calculating a tangent line at (x1, f(x1)) and finding the x-intercept 2. 1
1. Recall that the equation of a line with gradient m passing through the point (a, b) is y ẞ m(x-a). Show that if xn+1 is the x-intercept of the tangent to f at (xn, f(xn)), then Xn+1 Xn - f(xn) f'(xn) provided that f'(xn) ± 0. (1) [3 marks]
Transcribed Image Text:1. Recall that the equation of a line with gradient m passing through the point (a, b) is y ẞ m(x-a). Show that if xn+1 is the x-intercept of the tangent to f at (xn, f(xn)), then Xn+1 Xn - f(xn) f'(xn) provided that f'(xn) ± 0. (1) [3 marks]
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