Newton's method The following sequences come from the...

Newton's method The following sequences come from the recursion formula for Newton's method, $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ Do the sequences converge? If so, to what value? In each case, begin by identifying the function $f$ that generates the sequence. $$ \begin{array}{l}{\text { a. } x_{0}=1, \quad x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2 x_{n}}=\frac{x_{n}}{2}+\frac{1}{x_{n}}} \\ {\text { b. } x_{0}=1, \quad x_{n+1}=x_{n}-\frac{\tan x_{n}-1}{\sec ^{2} x_{n}}} \\ {\text { c. } x_{0}=1, \quad x_{n+1}=x_{n}-1}\end{array} $$

Newton's method The following sequences come from the
recursion formula for Newton's method,
$$
x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}
$$
Do the sequences converge? If so, to what value? In each case,
begin by identifying the function $f$ that generates the sequence.
$$
\begin{array}{l}{\text { a. } x_{0}=1, \quad x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2 x_{n}}=\frac{x_{n}}{2}+\frac{1}{x_{n}}} \\ {\text { b. } x_{0}=1, \quad x_{n+1}=x_{n}-\frac{\tan x_{n}-1}{\sec ^{2} x_{n}}} \\ {\text { c. } x_{0}=1, \quad x_{n+1}=x_{n}-1}\end{array}
$$

Key Concepts

Newton's Method

Newton's method is an iterative algorithm used to find the roots of a differentiable function. It works by linearizing the function at a current approximation point via its tangent line and then using the x-intercept of this tangent as the next approximation. This method is notable for its quadratic convergence under proper conditions, meaning that the error decreases rapidly once the iterates are close to the actual root.

Root-Finding Problems

A root-finding problem involves determining the value(s) of x for which a given function f(x) is zero. Newton's method serves as a powerful tool in addressing these problems by utilizing both the function and its derivative to iteratively approach the solution, making it particularly effective when the function behaves well near the desired root.

Convergence Analysis of Iterative Sequences

Convergence analysis is critical when using iterative methods like Newton's method to determine whether the sequence of approximations will settle at a fixed value (the root) or diverge away. This involves checking whether the conditions for rapid local (often quadratic) convergence are met and understanding how factors such as the choice of initial guess and the behavior of the derivative affect the overall stability of the method.

Function Identification in Newton's Method

Identifying the function corresponding to a given iteration is essential because the Newton iteration formula, x??? = x? - f(x?)/f'(x?), can be reversed engineered to deduce f and f'. Recognizing the underlying function helps in predicting which root the method is targeting and in assessing convergence properties based on the behavior of f() and its derivative.

Fixed Points and Stability in Iterative Methods

A fixed point in the context of iterative methods is a value that remains unchanged under the iteration process, meaning that it satisfies the condition x = x - f(x)/f'(x). In Newton’s method, fixed points correspond to the roots of the function. Studying the stability of these fixed points is important because stable (attractive) fixed points ensure that an iteration starting close enough will converge to the correct root, while unstable fixed points lead to divergence or erratic behavior.

Transcript

00:01 So newtons method helps us try to find the zeros of functions.

00:10 And we're given some of the sequences using newton's method.

00:18 And we want to figure out what these values approach, if they approach, and some function app that generates each of these.

00:28 So let's just go ahead and kind of start.

00:30 So here i have them kind of, on top of each other.

00:35 So notice this top function here could be f of x -in and if we take the derivative of f xn prime well that's going to be 2xn and then the derivative of 2 is just 0 so we get the same thing in the denominator so we can go ahead and assume that f x -n is equal to x squared n minus 2.

01:07 And since our initial guess here is supposed to be 1, so since we're assuming that x0 is supposed to be 1, we'll plug in this in for a -n, or x -n plus 1, x -n plus 1.

01:22 And i'm going to use this second equation to do this.

01:25 It'll give us 1 over 2 plus 1 over 2, which is 1.

01:32 And we can kind of tell that since we're adding xm plus 1 will always be, or actually this here should be x1, sorry about that.

01:49 So this is one, and we can kind of also further conclude that all of these will be bigger than or equal to at least zero since it is positive.

02:05 So what this tells us is we know that this is trying to get closer and closer to a zero of this function.

02:14 So if we were to factor this, we're going to get xn minus the square root of 2.

02:22 And let me use it a mile a little bit, and xn plus the square root of 2.

02:31 So if this is supposed to equal to 0, we would get xn is equal to the square root of 2.

02:37 Or xn is equal to the square root or the negative square root 2.

02:42 And since we're assuming all of these have to be bigger than n plus 1, that would tell us this here is what this should approach.

02:52 So we could just go ahead and make the assumption that the limit as in approach to infinity of xn is going to be the square root of 2.

03:03 And we can once again make this assumption just because we are, assuming that newton's method is actually going to give us a value that should be a zero of our function.

03:18 So that's what we end up getting for the first part.

03:23 Now for the second, well, we're going to follow pretty much the same steps that we did for.

03:32 Now, if we go ahead and let our numerator over here, the f of x.

03:41 So i'm just going to write f of x as opposed to f n or f x in and if we go ahead and take the derivative of that so x prime or the derivative of tangent is sequent squared x the derivative of negative 1 is 0 so the denominator is that so we can go ahead and make the assumption that f of x in this case is going to be tangent x minus 1 and so once again, newton's method, remember, we want to find zeros of our polynomial, or of our function.

04:27 So we can go ahead and set this equal to zero and then solve.

04:29 So we get tangent of x is equal to 1, and we know that x is supposed to be any value...

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