Use Newton's method [state the function f(x) you use] to estimate the given number. √(24)

Name: Problem 29, Chapter 3: Calculus: Early Transcendental Functions
Uploaded: 2020-06-13T01:26:09-08:00
Duration: 3 min 37 s
Channel: Grace Muhihu
Description: Use Newton's method [state the function f(x) you use] to estimate the given number. √(24)

Use Newton's method [state the function f(x) you use] to...

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Key Concepts

Newton's Method

Newton's method is an iterative procedure used to approximate the roots of a function. Starting from an initial guess, the method uses the formula x??? = x? - f(x?)/f'(x?) to generate progressively better approximations of the function's zero, making use of the function’s derivative.

Formulating the Function for Root-Finding

When solving problems using Newton's method, the equation of interest must be recast in the form f(x) = 0. For instance, to find an nth root of a number, one can define the function as f(x) = x? - a, thereby aligning the problem with the standard framework for Newton's method.

Derivative Calculation

The derivative f'(x) is central to applying Newton's method because it is used to calculate the slope of the tangent line at the current approximation. Accurate computation of the derivative is essential, as it directly influences the efficiency and convergence of the iterative process.

Iterative Convergence Criterion

Newton's method is an iterative process where the quality of approximation is improved step by step. Establishing a convergence criterion, such as a tolerance level for the difference between successive iterates or the value of f(x), is important to decide when the approximation is sufficiently close to the true root.

Transcript

00:01 Okay, so to determine this square root, what i'm going to do is i'm going to set this equal to x.

00:07 So x is equal to the square root of 23.

00:11 Then i'm going to square both sides, so x squared is equal to 23.

00:15 Then i'm going to bring 23 to the left hand side, which will give me x squared minus 23 is equal to 0.

00:22 So this is going to be my function.

00:24 So f of x is going to be equal to x squared minus 23.

00:29 I'm also going to take the derivative because i'm going to need that for my formulas later.

00:35 So f prime of x is equal to 2x.

00:38 Then for using newton's method to determine this, my formula that i'm going to use is x sub n plus 1 is equal to xn minus f of xn over f prime of xn.

01:00 Okay, so i want to use this.

01:02 So first i have to come up with a guess.

01:04 So i know that the square of 23 is in between the square of 16 and the square of 25.

01:09 So since i know it's now between 4 and 5, so i'm going to let my guess be 5.

01:14 So i'm going to set that equal to x0.

01:17 So x0 is equal to 5.

01:19 So to find x1, i'm going to use the formula that i wrote down here.

01:23 So x1 is equal to xn.

01:25 In this case, it's going to be x0, which is 5, minus f of x0...