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

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.

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.

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.

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.

Transcript

00:01 Okay, so to start to find x1, we're going to need our formula.

00:08 So x1 is equal to x0 minus f of x0 over f prime of x0.

00:22 Okay, so first we need to find f of x0.

00:26 So this is considered our f function.

00:30 So f of x0 means we plug negative 1 into here.

00:33 So with that we get negative 1 to the 4th minus 3 times negative 1 squared plus 1.

00:44 Okay, so if we do that, we get negative 1.

00:49 Okay, so then the next thing that we need to do is to find the derivative of this function here.

00:54 So f prime of x is equal to 4x cubed minus 6x.

01:00 Then from here we're going to plug in x0, which is negative 1.

01:03 So f prime of x0 is 4 times negative 1 cubed minus 6 times negative 1, which will give us positive 2.

01:17 Okay? so then we just need to plug this, this, and this into this formula.

01:23 So x1 is equal to x0, which is negative 1, minus f of x0, which is negative 1, over f prime of x0, which is 2.

01:35 So with that we get the x1 is equal to negative 1 half.

01:40 So right here the x1 is equal to negative 1 half.

01:48 Okay, so for x2, we're going to do the same thing.

01:52 Only our formula is going to look just a little bit different.

01:55 So x2 is x1 minus f of x1 over f prime of x1.

02:05 Okay, so here again, the first thing that we need to do is we need to find f of x1...

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