Sequences generated by Newton's method Newton's method,...

Sequences generated by Newton's method Newton's method, applied to a differentiable function $f(x)$ , begins with a starting value $x_{0}$ and constructs from it a sequence of numbers $\left\{x_{n}\right\}$ that under favorable circumstances converges to a zero of $f .$ The recursion formula for the sequence is $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ a. Show that the recursion formula for $f(x)=x^{2}-a, a>0$ can be written as $x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2 .$ b. Starting with $x_{0}=1$ and $a=3,$ calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Sequences generated by Newton's method Newton's method,
applied to a differentiable function $f(x)$ , begins with a starting
value $x_{0}$ and constructs from it a sequence of numbers $\left\{x_{n}\right\}$ that
under favorable circumstances converges to a zero of $f .$ The
recursion formula for the sequence is
$$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
a. Show that the recursion formula for $f(x)=x^{2}-a, a>0$
can be written as $x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2 .$
b. Starting with $x_{0}=1$ and $a=3,$ calculate successive terms
of the sequence until the display begins to repeat. What
number is being approximated? Explain.

Key Concepts

Newton's Method

Newton's Method is an iterative numerical technique used to approximate the roots of a differentiable function. The method starts with an initial guess and refines this guess by using the derivative of the function to move closer to a zero of the function with each iteration.

Recurrence Relation Transformation

Recurrence relations derived from Newton's Method can often be algebraically manipulated into simpler forms for specific functions. For example, when the function has the form x² - a, the general Newton's iteration can be rearranged into a form that directly averages the current guess and a/a value, streamlining computation.

Convergence of Iterative Sequences

The convergence of sequences in Newton's Method refers to the process by which repeated iterations yield values that increasingly approximate the actual root. Under favorable conditions, such as a good initial guess and appropriate function behavior, the sequence converges rapidly to the exact solution.

Square Root Approximation

Newton's Method is particularly useful in computing square roots by finding the positive root of the function x² - a. In such cases, the iterative formula can be transformed into a form that efficiently averages the guess with a divided value, providing a practical algorithm for approximating square roots.

Transcript

00:01 So here we're told that newton's method is this sequence given by xn plus 1 is equal to xn minus f of xin over f prime of xin.

00:17 And essentially what it does is it takes some initial starting value, which we call x not.

00:24 And if we have where f of x is differentiable and we have some other favorable conditions, then this.

00:33 Sequence we just listed should approach to a zero of our function f or in other words when we take that value and plug it into f we should get closer and closer to zero so at first what we want to show or we want to find that the recursion formula for f of x is equal to x squared minus a when a is strictly greater than zero and then given that x not is one and a is 3, we want to calculate successive terms of the sequence until it begins to just repeat or seems to repeat.

01:17 And what we want to do is decide what number is being approximated and we want to explain why.

01:25 So let's go ahead and first calculate what f prime of x should be.

01:32 So let's just go ahead and do that down here.

01:35 So f prime of x is going to be d by d x of x squared.

01:42 Minus a.

01:45 Well, the derivative of x squared is going to be 2x.

01:48 The derivative of a is zero since it's a constant.

01:52 Now let's go ahead and plug f of x and f prime of x into our recursion formula up here.

02:03 So we get x -n minus, well f of x was x squared minus a.

02:11 And then we found here that f -prime of x was 2 over x.

02:17 Now we can go ahead and divide x into here, and we'll get, oh, this should be xn, sorry.

02:28 So xn minus, and now dividing the xn into everything, we'd get one half xn minus a over xn, like that.

02:46 So this is what they wanted us to show in part a.

02:54 So that's the end of part a.

02:59 And then, after we've shown that, we want to start with x not equal to 1 and a is equal to 3 and calculate a few of the terms.

03:14 All right, so i'm just going to go ahead and maybe just do it for the first two terms.

03:18 And then on the second page, i'll have some values for us using an excel spreadsheet...

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