Suppose we with to develop an iterative method to compute the square root of a g
ID: 2945411 • Letter: S
Question
Suppose we with to develop an iterative method to compute the square root of a given positive number y, i.e., to solve the nonlinear equation f(x) = x2 - y = 0 given the value of y. Each of the functions g1, and g2 listed below gives a fixed-point problem that is equivalent to the equation f(x) = 0. For each of these functions, determine whether the corresponding fixed-point iteration scheme xk + 1 = gk(xk) locally convergent to if y = 3. Explain your reasoning in each case. g2(x) = 1 + x - x2/y. What is the fixed-point iteration function, Phi (x) given by the Newton - Raphson method for this particular problem?Explanation / Answer
a)As long as the absolute value of the derivative is less than 1 (this gives us the ratio), it will converge.
The derivative here is 1 - 2x/3. Evaluating this at x=3, we get 1 - 1.154701 = - .154701, so it converges.
b)The Newton-Raphson method is x - f(x)/f'(x)
We are trying to find the 0 of y = x2 - 3
f'(x) = 2x
Then, our equation is
g(x) = x-f(x)/f'(x) = x - (x2 - 3)/2x = x - (x/2 - 3/2x) = x/2 + 3/2x = 1/2(x + 3/x) (I put it in this form, because this is the traditional form that we see this equation.)
The derivative of this function 1/2(x + 3/x) = 1/2 - 3/(2x2) If we plug in x = 3 here, we get 1/2 - 3/(2*3) = 0. Thus, this method also converges. In fact, the 0 tells us that it will converge much more quickly than the method in a. This 0 is not an accident, but a function of the method. We will always get this result for this method.