![]() ![]() This type of process, where each x n x n is defined in terms of x n − 1 x n − 1 by repeating the same function, is an example of an iterative process. In particular, by defining the function F ( x ) = x −, F ( x ) = x −, we can rewrite Equation 4.8 as x n = F ( x n − 1 ). When using Newton’s method, each approximation after the initial guess is defined in terms of the previous approximation by using the same formula. Use Newton’s method to approximate 3 3 by letting f ( x ) = x 2 − 3 f ( x ) = x 2 − 3 and x 0 = 3. Typically, the numbers x 0, x 1, x 2 ,… x 0, x 1, x 2 ,… quickly approach an actual root x *, x *, as shown in the following figure. We continue in this way, deriving a list of approximations: x 0, x 1, x 2 ,…. If f ′ ( x 1 ) ≠ 0, f ′ ( x 1 ) ≠ 0, this tangent line also intersects the x x-axis, producing another approximation, x 2. Next we draw the tangent line to f f at x 1. Typically, x 1 x 1 is closer than x 0 x 0 to an actual root. Now let x 1 x 1 be the next approximation to the actual root. If f ′ ( x 0 ) ≠ 0, f ′ ( x 0 ) ≠ 0, this tangent line intersects the x x-axis at some point ( x 1, 0 ). We then draw the tangent line to f f at x 0. By sketching a graph of f, f, we can estimate a root of f ( x ) = 0. Newton’s method makes use of the following idea to approximate the solutions of f ( x ) = 0. In cases such as these, we can use Newton’s method to approximate the roots. No simple formula exists for the solutions of this equation. For example, consider the task of finding solutions of tan ( x ) − x = 0. Similar difficulties exist for nonpolynomial functions. No formula exists that allows us to find the solutions of f ( x ) = 0. For example, consider the functionį ( x ) = x 5 + 8 x 4 + 4 x 3 − 2 x − 7. Also, if f f is a polynomial of degree 5 5 or greater, it is known that no such formulas exist. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. However, for polynomials of degree 3 3 or more, finding roots of f f becomes more complicated. If f f is the second-degree polynomial f ( x ) = a x 2 + b x + c, f ( x ) = a x 2 + b x + c, the solutions of f ( x ) = 0 f ( x ) = 0 can be found by using the quadratic formula. If f f is the first-degree polynomial f ( x ) = a x + b, f ( x ) = a x + b, then the solution of f ( x ) = 0 f ( x ) = 0 is given by the formula x = − b a. Describing Newton’s MethodĬonsider the task of finding the solutions of f ( x ) = 0. This technique makes use of tangent line approximations and is behind the method used often by calculators and computers to find zeroes. In this section, we take a look at a technique that provides a very efficient way of approximating the zeroes of functions. For most functions, however, it is difficult-if not impossible-to calculate their zeroes explicitly. In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form f ( x ) = 0. 4.9.4 Apply iterative processes to various situations.4.9.3 Recognize when Newton’s method does not work.4.9.2 Explain what an iterative process means.4.9.1 Describe the steps of Newton’s method.Rw=1 % I was thinking I should make the Residual 1 for the next time step. If Rw(m+1)>10^-6 %Check on the level of convergence Thank you! while Rw(m)>10^-6 % Convergence condition I've been trying to figure this out for a while so any help will be appreciated. I'm confused on how the time stepping comes in. The problem I am having is how to implement the time step dt as t=0:Tmax/dt where Tmax is say 10. I know to check the convergence of the solution as the iterations go on. ![]() ![]() After that I know to add the value to w^m and get an update value for w at the next m iteration. I understand that with each iteration, one makes an initial guess, calculates the residual and solve for the change. I'm trying to implement a backward Euler Scheme using the Newton Raphson iteration.
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