Slow convergence Consider the function f(x)=x^5+4 x^4+x^3-10...

Key Concepts

Multiple Roots

When a function has multiple (repeated) roots, Newton's method can suffer from slow convergence despite having a theoretically quadratic convergence rate for simple roots. This slowdown occurs because, at a repeated root, the derivative of the function is close to zero or less informative, which effectively reduces the order of convergence. Special modifications or alternative methods are often considered to efficiently handle cases involving multiple roots.

Choice of Initial Approximation

The initial estimate in Newton's method critically influences the outcome and efficiency of the convergence process. Depending on the starting guess, the sequence of approximations may converge to different roots, may converge slowly, or may not converge at all. The sensitivity to initial approximations underlines the importance of choosing a starting value within the basin of attraction of the desired root.

Newton's Method

Newton's method is an iterative numerical technique used to find roots of a function by approximating the function locally with its tangent lines. At each iteration, the next approximation is computed by subtracting the ratio of the function value to its derivative from the current estimate. This method is particularly powerful because it generally exhibits quadratic convergence near simple roots under favorable conditions.

Convergence Behavior

The convergence of an iterative method like Newton's method refers to how quickly the sequence of approximations approaches the true root. For simple roots, one typically observes quadratic convergence, meaning the number of correct digits roughly doubles with each iteration. However, when convergence is slow, it may indicate that the underlying assumptions (such as a non-zero derivative) are not fully met, or that the behavior of the function near the root leads to a deceleration in the rate at which the approximations improve.

Transcript

00:01 Okay, so we need to use new news method to see how the initial approximations of negative 1, negative 0 .2, positive 0 .2, and 2 fair when it comes to approaching the actual answers of 1 and negative 2.

00:17 So before we start, i'm going to post a picture of the graph here like that.

00:26 Let me move to the right of it.

00:28 And as you can see, it does indeed have solutions at negative 2 and 1.

00:34 For x is equal to negative 1, that's going to be around here.

00:38 At x is equal to negative 0 .2, it's going to be right here at the peak, and that's going to be important in a reason i'll get to in a bit.

00:47 At x is equal to positive 0 .2, it's somewhere around here.

00:52 And at x equals 2, which is technically not on the graph, it's very far up.

00:59 So let's see how these do when it comes to conversion towards the answer.

01:05 So now we need to find the formula for our x of n plus 1.

01:12 And actually, as it turns out, f of x is factorizable as x minus 1 squared times x plus 2, cube.

01:27 And after some algebra, we can see that f prime of x is equal to x plus 1.

01:34 Times x plus 2 squared times 5x plus 1.

01:46 That means our differential term of f of x divided by f prime of x is equal to x minus 1 times x plus 2 divided by 5x plus 1.

02:03 That means our x minus f of x divided by f prime of x is equal to 4x squared plus 2 divided by 5x plus 1 and that's what we're going to plug into our x over n plus 1.

02:22 So we say this is equal to 4 times c2 squared plus 2 divided by 5 times c2 plus 1...

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