The functions f(x)=(x-1)^2 and g(x)=x^2-1 both have a root...

The functions $f(x)=(x-1)^{2}$ and $g(x)=x^{2}-1$ both have a root at $x=1 .$ Apply Newton's method to both functions with an initial approximation $x_{0}=2 .$ Compare the rates at which the method converges in these cases, and give an explanation.

Key Concepts

Newton's Method

Newton's method is an iterative numerical technique for finding successively better approximations to the roots (or zeros) of a real-valued function. It uses the formula x??? = x? - f(x?)/f?(x?), which requires both the function and its derivative. This method's effectiveness and speed of convergence depend heavily on the properties of the function near the root.

Rate of Convergence

The convergence rate of an iterative method describes how quickly the sequence of approximations approaches the true solution. Under favorable conditions, Newton's method typically converges quadratically, meaning that the number of correct digits approximately doubles with each iteration. However, this rapid convergence is contingent upon the root being simple and the derivative not vanishing at the root.

Multiplicity of Roots

Root multiplicity refers to the number of times a particular root appears in the factorization of a polynomial. When a function has a repeated (multiple) root, its derivative is zero at that root, undermining one of Newton's method’s key assumptions. This can lead to significantly slower convergence rates, often reducing the quadratic convergence to linear, because the function behaves nearly flat near the multiple root.

Transcript

00:01 Problem number 60, to approximate the root r of the function fn, the initial approximation x note, we use the newton's method.

00:11 The formula that we use for this is x n plus 1 equals x n minus f of x n over f dash of xn.

00:23 Part number a of the problem.

00:25 In our case, f of x equals x minus 1, all power 2, and hence f dash of x equals 4.

00:33 X minus 1 which is equal to 2x minus 2 and we will use a value of x node that's equal to 2 we should stop calculating approximations when two successive approximations agree to 5 digits to the right of the decimal point after rounding okay since i've already solved this problem i know that it's it's going to be a lot of iterations so this is why i'm making the table the columns so long we're not going to solve each iteration step by step, but i'm going to go through most of them.

01:38 So the first iteration, initial iteration, with an x note of 2, we get a value 1, 2, 1 .5, 2, 2 .5, 2 .5.

01:52 0 .5, 1 .15.

01:56 Still not the same, so we go for another iteration.

01:59 Sorry i my bed this was supposed to be 1 .25 okay so since these two values are still not the same we will have to go for another iteration but i'm going to skip skip these steps and go on to the final four three iterations okay so this will go on to till the 15th iteration should have a value of 1 .004, 1 .6 multiply 10 power negative 9, 0 .0 ,0008, 1 .00002, 16th iteration, 1 .002.

03:18 In the previous value in the 14th iteration we had a value of 1 .00004, so this is why we had to move on to the 16th iteration of 0 .00, sorry, 0, 0 2, 4 multiplied 10 to the power of negative 10, 0 .004, 1 .00001.

03:49 Still we have to go for another iteration...

Please give Ace some feedback