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Newton's Method

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Newton's Method 0:45
    • Newton's Method
  • Example I: Find x2 and x3 13:18
  • Example II: Use Newton's Method to Approximate 15:48
  • Example III: Find the Root of the Following Equation to 6 Decimal Places 19:57
  • Example IV: Use Newton's Method to Find the Coordinates of the Inflection Point 23:11

Transcription: Newton's Method

Hello, and welcome back to, welcome back to AP Calculus.0000

Today, we are going to be talking about something called Newton's method.0004

It is a way of finding the roots of any function, the 0’s if you will, places where it touches the x axis.0007

Interestingly enough, for all the powerful methods that we have developed over the years,0018

the numerical methods for finding roots, Newton's method still actually ends up being the one that most calculators0023

and computers used to do so because it finds them very quickly.0030

We use the term converge, it converges to the root very quickly.0035

Let us jump on in and see if we can make some sense of this.0039

It is actually very simple.0042

Let us start off with, there is a way of using the derivative, since we are still talking about applications of derivatives, 0048

there is a way of using the derivative to find the root also called the 0’s of a function.0067

Here is the geometry of it.0085

I’m going to go ahead and draw a little like this and I will go like that.0086

It involves making it so we see that the root is like right here.0096

We want to find what that root is and this is just some f(x).0102

Let us start by making a guess, either by looking at the graph or maybe you know something about the function, just making a first guess.0109

Let us guess like right over here.0118

We will call that our x1, our first guess.0121

If we go up to where it meets the graph, this is going to be the point x.0125

This is going to be the point x1 f(x1).0136

I will go ahead and put brackets around that since I got parentheses around there.0143

That is that one.0148

At this point, I take the derivative of that point and find the tangent line.0150

If I follow that tangent line to where it touches the x axis, not the best tangent line but you get the point, that is going to be our x2.0157

This x2, the point is x2, let me write it over here.0174

The coordinate of that point is x sub 2,0.0184

Starting with x1, I go up to the graph.0190

I follow that tangent line, the derivative, the f’, down to where it hits here.0193

Let us find the equation for this line.0200

Let me actually write down the process.0205

Make a first guess, this is our x1.0208

Follow f’(x), the slope, follow f’(x) down to where it touches the x axis.0216

This is going to be our x2,0.0239

The equation of the line, the equation of this line is just y - y1 = m × x - x1.0243

We know that already.0261

This is our x1 y1 x2 y2, we can go ahead and write, 0266

I will take this, I will write y, f(x1) - 0 = f’(x) × x -, I will do x1 – x2.0276

x1 – x2, when I rearrange this, I get x2 is equal to x1 – f(x1) divided by f’(x1).0312

Having picked a first guess, x1, this formula gives me a method of finding x2.0335

If I just keep going, x1 follow down to here.0343

Notice x2 is actually closer to my root.0347

From x2, I go up and I do the process again.0350

Now the tangent line is going to put me here at x3.0352

If I go up again, it is going to put me at x4, x5, x6.0357

Within about 4 or 5, you are actually going to hit the root to a very high degree of accuracy.0363

That is all we are doing, this formula, once I pick my x1, it gives me way of finding my x2.0368

Then, my x2 becomes my x1.0376

And I do it again, I find my x3, so on and so forth.0378

I just keep following the derivative until I actually touch my root or get really close to it.0382

That is the whole idea behind Newton's method.0389

Let me write all of this down here.0399

Having chosen an x1, we have a method for finding x2 which is closer to the root, and then, we just repeat.0401

In other words, x3 = x2 – f(x2)/ f’(x2), and so on.0435

And so on until x sub n + 1 is equal to x sub n.0450

When you take, and let us say an x5, and then you end up doing an x6,0456

if x6 = x5 to the number of decimal degrees of accuracy that you like, that you are happy with, you can stop.0461

That is your root, that is all we are doing.0469

The general formula for Newton's method is x ⁺n + 1 = x ⁺n - f(x) ⁺n divided by f’ (x) ⁺n.0471

If the sequence x sub 1, x sub 2, x sub 3, and so on, if the sequence actually converges,0493

actually gets close to a number, that number is a root of f(x).0507

Very powerful, actually.0528

The only thing you have to be careful of is this method does not always work.0531

We will show you how this method does not always work but there are ways around it.0537

The way around it is to actually either make a better guess, something closer,0545

and that comes from knowing the function, graph the function.0550

This method does not always work, we just have to watch out for that.0555

Let us show you the circumstances under which this might actually fail.0561

Let me go to the next page and do this actually.0567

I have some function like that so let us say I take my x1 over here.0575

I’m looking for, this is my root right there.0583

Let us say I take my x1 over here.0586

I go up to where it touches the graph and I follow.0588

Notice it puts me like, let us say it is over here.0596

This is my x2, now I take from my x2, I go down and I follow the tangent line.0602

It puts me at x3.0615

Depending on where you take your x1, where you get your first slope,0617

it might actually end up shooting you farther out, when you come around.0622

x2 might be farther out, then x3 might be farther up, and your x1 and x2, and so on, and it will just end up diverging.0626

It will not actually get close to a number, those just blow up.0635

The solution to that is just basically choose a better x.0638

You do that by basically graphing the function, looking at a function and making a better guess.0643

As long as you stay reasonably close, you will actually converge to the root.0649

Here, the sequence starts to diverge.0656

We really have to choose wisely.0672

Again, this is just one more tool in your toolbox.0683

That is all, that is all this is, knowing what the graph looks like goes a long way.0696

Again, that is what we have been doing for so long.0716

We have been graphing functions, we have been taking derivatives, finding critical points, maximums, minimums.0717

We have used all of this information to graph the function.0722

We want to use the graph of the function to help us make a good guess, as to where the root might be.0726

Last of all, before we start the examples.0733

This technique, clearly, because you have the x ⁺n + 1 = x ⁺n – f(x) ⁺n/ f’ x ⁺n.0735

It lends itself very well in the computation.0747

In fact, this is the algorithm that your calculator is using to find roots of the equations.0750

This technique lends itself.0755

To programmable and computational devices because it is algorithmic.0765

It follows a series of steps, algorithmic, a series of reputable steps.0778

That is really all an algorithm is.0790

Let us do some examples.0797

The following function and initial guess x1, find x2 and x3.0802

I do not even have a function here.0811

I will tell you what the function is.0813

F(x) is x⁵ - 15x + 10 and our x1 is going to be 1.0815

Very simple, our x1 is 1, we know our x2, it is just x1 – f(x1)/ f’(x1).0829

Our f’ is nothing more than 5x⁴ – 15.0846

This is going to equal 1 – f(1)/ f’(1), that equals 1 - f(1) is going to b -4/-10.0859

That is going to be 0.6.0885

Therefore, this is our x2.0891

Therefore, our x3 is our x2 which is 0.6 - f(0.6)/ f’(0.6).0894

When you do that calculation, you get 0.675, that is x3.0908

If you want to see what x4 looks like, just keep going.0915

X4 is equal to x3 which is 0.675 – f(0.675)/ f’(0.675).0918

When you do this, you end up with 0.6761.0930

Clearly, it converged very quickly, 0.6761, 0.675.0937

Very good, this is a fantastic algorithm.0944

Use Newton’s method to approximate the 6√27 to 8 decimal places.0950

When you are given a number to approximate, treat it like this.0957

x = 6√27, that is equal to 27¹/6.0965

That is the same as x⁶ = 27, x⁶ - 27 = 0.0982

Newton's method works because we are looking for roots, we are looking for 0’s,0991

where it hits the x axis, which means we have to have some function set equal to 0.0997

That is the whole idea behind Newton’s method.1002

Now choose x1, choose an x sub 1.1005

How do we choose? Let us try some things.1013

If I take x sub 1 is equal to 1, 1⁶ is equal to 1.1014

Let us try something else, let us try 1.5.1021

If I take 1.5⁶, that is going to be 11.4.1024

Let us go a little higher, let us try 1.7.1030

We are trying to choose an x sub 1, a first guess.1034

When I take 1.7 and I raise it to the 6th power, I get 24.1039

24 is close to 27, let us go ahead and start with that.1044

Let x sub 1 equal 1.7.1050

x sub 2 is equal to 1.7 - f(1.7) divided by f’(1.7).1055

This is f, f’ is just 6x⁵, that is it, that is all you are doing.1068

F is on top, f’ is on the bottom.1083

This is 1.7, by the way.1085

When I solve that, I get 1.7336, now I find x3.1090

x3 is equal to this, 1.7336 – f(1.7336) divided by f’(1.7336 ).1098

I put 1.7336 into f, I get a number.1113

I take the 1.7336 into f’, I get a number.1117

I divide the f(1.7336) divided by the f’(1.7336) and I subtract it from 1.7336.1121

When I do that, I get 1.732054.1131

I do the same thing, x sub 4 = now this, 1.732054 – f 1.732054 divided by f’ 1.732054.1140

I end up with 1.7320508.1163

That is fine, does not really matter.1176

Clearly, you just keep going until you get, in this case, a decimal place that match.1178

If you end up doing x5, x6, or x7, and you end up getting the same answer as you did before, 1183

to 8 decimal places that is where you can stop.1188

Very simple, nice and straight algorithmic process.1191

Find the root of the following equation to 6 decimal places.1198

This is our f(x), our f(x) is equal to sin x + cos x – x³, because we want an f(x) to set equal to 0.1203

Just bring this over to that side so that you have 0 on the right side.1222

This is our f(x) right here.1228

Now f’(x) = cos x – sin x - 3x².1232

When we look at the graph of this function to decide where our x1 is going to be, 1247

when we look at a graph of that function to see what we might choose for x sub 1, we choose x = 1.1.1254

I look at a graph of this function, I just graph it using my graphing utility, whatever one I like.1293

I just see where it crosses the x axis and I just estimate, something simple, 1.1.1299

We go through our process.1308

x sub 2 = 1.1 - f(1.1) divided by f’(1.1).1311

When I do that, I get 1.103394.1323

x3 is equal to 1.103394 – f(1.103394) divided by f’(1.103394).1330

I get 1.103382.1352

When I do x4, I end up getting 1.103382.1358

This and this match to 6 decimal places, I can stop, that is my root.1366

That is extraordinary, 6 decimal places is extraordinary.1371

2 decimal places is extraordinary.1374

Anything more than that is a bonus.1376

I only went through four iterations, three iterations, 1, 2, 3, absolutely fantastic algorithm.1379

Let us see what is next here.1387

Use Newton's method to find the coordinates of the inflection point for the following function.1392

F(x) = cos e ⁺x of the interval from 0 to π/3.1396

Here, f(x) = cos(e ⁺x).1404

F’(x) = -sin e ⁺x × e ⁺x.1416

F” is equal to –sin e ⁺x × e ⁺x, this is product rule, + e ⁺x × cos(e ⁺x ) × e ⁺x.1428

A point of inflection is where the second derivative f’ is equal to 0.1450

F”(x) = 0, gives a point of inflection.1456

We want the root of f”(x) not f(x).1474

Be very clear, they are asking for us to find the coordinates of inflection point.1487

We get an inflection point from the second derivative.1492

We want the roots of f’(x).1495

In other words, this is our f(x).1499

Our f(x) is f”(x), we want to roots of that.1510

That is going to give us an inflection point.1514

Because of Newton's method, the function that we are looking for, we also need its derivative.1522

We need F’(x) which is actually f’’’.1527

One more derivative, let us go ahead and do,1544

We said that f(x) is equal to -sin e ⁺x × e ⁺x + e ⁺2x × cos(e ⁺x).1549

F’(x) is going to equal –sin(e ⁺x) e ⁺x + e ⁺x cos e ⁺x e ⁺x 1569

+ e ⁺2x × -sin e ⁺x × e ⁺x + cos(e ⁺x) × e ⁺x.1588

I have my f and I have my f’.1606

We form our x ⁺n + 1 = x ⁺n – f(x) ⁺n divided by f’(x) ⁺n.1622

I’m hoping that you are doing this by using your calculator tool, that gives you a table of data and things like that.1638

Looking at a graph, we have to choose our x sub 1.1643

Looking at a graph of f(x), the original function which was the cos(e ⁺x) between 0 and π/3.1650

When I look at the graph, an inflection point appears to be around x = 0.8.1671

I’m going to take 0.8 as my x sub 1.1688

x sub 1 = 0.8, x sub 2, I use this formula.1693

I end up with 0.721951.1700

x sub 3, I use this formula, I get 0.707864.1708

x sub 4 = 0.707424.1719

x sub 5 = 0.707424.1727

It ends up being the same as the thing before, to 6 decimal places.1733

I go ahead and I stop there.1735

They wanted the coordinates, in other words they wanted both x and y.1737

What I found was just the x, let us go ahead and find y.1742

F(0.707424), the original function was -0.442121.1753

Therefore, our coordinate equals 0.707424 - 0.442121.1768

Let us see if what a graph of this looks like.1794

Here is the graph, 0 π/6 roughly here.1798

An inflection point, concave down.1803

And it looks like here, concave up, there is your point 0.707 - 0.4, whatever.1805

There you go, thank you so much for joining us here at

We will see you next time, bye.1821