Use Newton's method with the given x0 to (a) compute x1 and...

Use Newton's method with the given $x_{0}$ to
(a) compute $x_{1}$ and $x_{2}$ by hand and (b) use a computer or calculator to find the root to at least five decimal places of accuracy.
$$x^{4}-3 x^{2}+1=0, x_{0}=1$$

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Key Concepts

Newton's Method

Newton's method is a numerical technique used to approximate the roots of a real-valued function. It involves iteratively applying the formula x(n+1) = x(n) - f(x(n))/f'(x(n)), which uses the function and its derivative to find successively closer approximations of the true root starting from an initial guess.

Derivative

In the context of Newton's method, the derivative of the function provides the slope of the tangent at a given point. This slope is used to determine the adjustment needed to move the current approximation closer to the root, making the derivative a critical component for the method's accuracy and convergence.

Iterative Methods

Iterative methods are approaches that generate a sequence of approximations converging to an exact solution. In problems like this, the method is applied repeatedly (iteration by iteration) until the sequence of values approaches the actual root within a specified level of accuracy.

Convergence and Accuracy

Convergence in root-finding methods refers to the process by which the iterative approximations approach the true value of the root. Ensuring accuracy, such as obtaining five decimal places of precision, involves assessing that the change between successive iterations is below a predefined tolerance level.

Transcript

00:01 Okay, so to find x1, we're going to use the formula.

00:07 X1 is equal to x0 minus f of x0 over f of x0.

00:20 Okay, so the first thing that we're going to need to find in order to complete this formula, and to solve x1 is to find f of x0.

00:31 So with that, we're going to plug one into this function.

00:35 So that's going to be 1 to the 4th minus 3 times 1 squared plus 1.

00:46 And with that, we will get negative 1.

00:50 Now if we take the derivative of this function, so f prime of x, so that'll be 4x cubed minus 6x.

01:00 So then we can plug in this x0, which is 1, into this derivative.

01:09 So if we do that, we get 4 times 1, cube minus 6 times 1, which will give us 4 minus 6, which is equal to negative 2.

01:20 Okay, so we're going to plug in this, this, and this into this formula to get x1.

01:26 So x1 is going to be equal to x0 minus f of x0, which we figured out up here to be negative 1 over f prime of x0, which is negative 2.

01:39 And with that, we'll get positive one -half.

01:43 So x -1 is going to be equal to one -half.

01:48 Okay, so we're going to do the same thing to find x -2, except our formula is going to be just a little bit different.

01:55 X -2 is equal to x -1 minus f -of -x1 over f -prime of x -1.

02:05 So that means that we need to plug in x -1, which is one -half, and two...

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