For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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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

### AP Calculus AB Online Prep Course

### Transcription: Newton's Method

*Hello, and welcome back to www.educator.com, 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 calculators*0023

*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 www.educator.com.*1818

*We will see you next time, bye.*1821

2 answers

Last reply by: Neeraj Lalwani

Tue Jul 11, 2017 6:56 PM

Post by Neeraj Lalwani on July 8 at 12:06:01 AM

Hello professor,

I sincerely thank you for your amazing lectures. However, I have an important question: Since you mentioned that graphing calculators use this method as their algorithm, then why do you use a graphing calculator to approximate the root? Sorry, but this paradox confuses me; can you please explain this phenomenon?

Thank you.