Use Newton\'s Method to estimate the roots of f(x)=x^2-2 Compute the exact solut

Question

Use Newton's Method to estimate the roots of f(x)=x^2-2 Compute the exact solution via the quadratic formula. Use x_0=1 and perform 2 Newton iterations. Sketch and label your iterates on the real number line. doubleheadarrow Does it look like the process will converge? Use z0=i, and perform 2 Newton iterations. (Use a calculator for this--Google can do it! Or here is a good online complex calculator. The computations are pretty ugly otherwise. Simplify each iteration to a+bi form.) Does it look like the process will converge?

Explanation / Answer

Here, the question is related to Newton's method which can easily be done by understanding the same.

Let A > 0 be a positive real number. We want to show that there is a real number x with x 2 = A. We already know that for many real numbers, such as A = 2, there is no rational number x with this property. Formally, let fx) := x 2 A. We want to solve the equation f(x) = 0. Newton gave a useful general recipe for solving equations of the form f(x) = 0. Say we have some approximation xk to a solution. He showed how to get a better approximation xk+1 . It works most of the time if your approximation is close enough to the solution. Here’s the procedure. Go to the point (xk, f(xk)) and find the tangent line. Its equation is y = f(xk) + f (xk)(x xk). The next approximation, xk+1 , is where this tangent line crosses the x axis. Thus, 0 = f(xk) + f (xk)(xk+1 xk), that is, xk+1 = xk f(xk) f (xk) . Applied to compute square roots, so f(x) := x 2 A, this gives xk+1 = 1 2 xk + A xk . (1) From this, by simple algebra we find that xk+1 xk = 1 2xk (A x 2 k ). (2) Pick some x0 so that x 2 0 > A. then equation (2) above shows that subsequent approximations x1 , x2 , . . . , are monotone decreasing. Equation (2) then shows that the sequence x1 x2 x3 ..., is monotone decreasing and non-negative. By the monotone convergence property, it thus converges to some limit x. I claim that x 2 = A. Rewrite (2) as A x 2 k = 2xk(xk+1 xk) and let k . Since xk+1 xk 0 and xk is bounded, this is obvious. We now know that A exists as a real number. then it is simple to use (1) to verify that xk+1 A = 1 2xk (xk A) 2 . (3) Equation (3) measures the error xk+1 A. It shows that the error at the next step is the square of the error in the previous step. Thus, if the error at some step is roughly 106 (so 6 decimal places), then at the next step the error is roughly 10 12 (so 12 decimal places).

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