A program for using Newton's method on a grapher Let f(x)=x^3+3 x+1 . Here is a home screen prog

A program for using Newton's method on a grapher Let...

Key Concepts

Newton's Method

Newton's Method is an iterative numerical technique used to approximate the roots of a differentiable function. The method works by starting with an initial guess and then repeatedly applying the formula x??? = x? - f(x?)/f'(x?) to converge to a solution where f(x) equals zero.

Derivative

The derivative of a function represents the instantaneous rate of change and is critical in Newton's Method. It is used to calculate the slope of the tangent line at the current approximation, which guides the next estimate of the root.

Initial Approximation

The initial approximation is the starting value for the iterative process in Newton's Method. Its proximity to the actual root significantly influences the convergence speed and success of the method, making a good initial guess essential.

Iterative Process

The iterative process in Newton's Method involves applying the update formula repeatedly. Each iteration uses the current approximation and its derivative to generate a new, typically improved, estimate of the root until convergence criteria are met.

Convergence

Convergence in the context of Newton's Method refers to the process by which the sequence of approximations approaches the true root of the function. The method converges rapidly under favorable conditions; however, convergence is not guaranteed for every function or initial guess.

Transcript

00:01 What we want to do is to walk through actually a computer calculator process in order to use newton's method to approximate solutions, root zeros.

00:13 And so you know this newton's method could be very tedious.

00:16 So there's actually a calculator stepage that we could do on the ti 83, 84, or 89, or probably any graphing calculator to help alleviate this process.

00:30 And so what we're going to do is we're going to look at the function.

00:36 F of x is equal to x cubed plus 3x plus 1.

00:43 And what we're going to do is i'm going to use a ti 84.

00:47 We are actually going to put this in y1.

00:51 And then we're going to put the derivative, which is 3x squared plus 3x, not 3x.

01:01 Plus three, we're going to actually put that into y2 of our graphing calculator.

01:11 And then what we're also going to do is we're going to actually be able to kind of store some things in some features.

01:20 So what i'm going to do is go ahead and pull up my ti -84 graphing calculator.

01:29 And what i'm going to do is turn it on.

01:32 And then in y equals, i'm actually going to make sure i'm going to put in, you know, my x cubed into y1.

01:53 And then in y2, i'm actually going to put in 3x squared plus 3.

02:04 And i'm not going to graph it at all.

02:06 I'm just going to have it residing in y1 and y2.

02:10 And then i'm actually going to go back to my home screen.

02:16 And what i'm going to do is i am actually going to store in.

02:21 We are going to be looking at x equal to negative point three.

02:27 And so what i'm going to do is store.

02:32 Point three, i'm actually going to store it.

02:35 And you notice that the store feature is, actually at right here.

02:44 So i'm going to hit that and i'm going to store that into x.

02:49 And then what i'm going to do, and you notice i get negative 0 .3 when i store it into x.

02:55 Now what i'm going to do is i'm going to actually store x minus and then because this would be, because our newton's method, remember our newton's method is x of n plus 1 is equal to x of n, minus the function over its derivative evaluated x of n.

03:16 So that's kind of what i'm going to be doing is i'm going to actually go to vars and then yvars and function.

03:25 And i'm going to do y1.

03:27 And then i'm going to divide that by the derivative, which is stored in y2.

03:38 And there i have it.

03:39 And then i'm going to store this into x.

03:44 And so as i hit that, i get negative 0 .3 for the first time.

03:54 And then if i keep hitting enter, i notice i get now negative 3 .22.

04:02 And then i'm going to do it until i get no difference.

04:06 And you notice the third time i hit it, i get relatively no difference between when i hit it the second time and when i hit it the third time.

04:17 So i get negative 3 .2218.

04:21 So if i look at the 5245, we get 5.

04:27 I can hit it one more time.

04:29 And once again, there's absolutely no difference.

04:32 And so that is telling me that the solution to that function or where it crosses the x -axis, which is also known as a zero or a root, is actually equal to x.

04:48 Equal so i'm going to say x equal and of course my calculator went away negative point 322 pull this up again um 1 9 so that's 1 9 and that's where it crosses x -axis okay and so now what i want to do is go ahead and validate that with using the graph.

05:28 So i'm actually going to go ahead and go back and use the graph.

05:34 And i'm not going to, i'm not going to go ahead and use the derivative.

05:40 So i'm going to take that away now.

05:43 And i'm going to clear that.

05:48 I'm going to clear that out.

05:49 I'm going to clear this out.

05:51 And i'm going to actually just graph the x cubed.

05:54 And i'm going to probably have to zoom six.

06:00 And you notice it crosses the x -axis right here.

06:03 So i'm actually going to zoom box it in.

06:16 I hit enter and then i'm a zoom box to really zoom in on where that crosses.

06:27 And now i'm going to use my features in my calculator, which actually does the newton's method itself to calculate where it crosses.

06:35 So i'm going to actually do second calculate.

06:38 I'm a calculator.

06:41 And so remember we do a left bank, a right bound and then actually where i think it is and sure enough that tells me at negative 0 .32219.

07:00 So it validates what i was doing.

07:03 Okay.

07:04 So graphically from calculator we get x equal to the same thing, negative 0 .3 -2219.

07:23 So that is a a root or a zero or a solution for that.

07:29 Now, i can also do it if i wanted to, we started with x0 equal to negative 0 .3 to begin with.

07:39 Okay.

07:40 So let's say i didn't know exactly.

07:43 And so we weren't told that.

07:45 So what happens if i pick another x value? so let's say i do x0 is equal to, so let's say i know from the intermediate value theorem, value theorem, that f of 0 is 1, and f of negative 1 is negative 3.

08:18 So that tells me somewhere between 0 and negative 1 is where it crosses the x -axis...

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