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Lecture Comments (22)

1 answer

Last reply by: Dr. William Murray
Thu Aug 4, 2016 6:04 PM

Post by Peter Ke on July 31, 2016

For example 1, at 7:10 why did you compare the fraction to 1/100?
I am so lost here.

Please explain.

1 answer

Last reply by: Dr. William Murray
Mon Jun 13, 2016 9:07 PM

Post by Silvia Gonzalez on June 10, 2016

Thank you Professor Murray, I have enjoyed this course a lot. I have a question about this lecture. In examples 1 and 4, the wording of the problem asks about the Taylor polynomial for cos x, however in both solutions you use the Maclaurin expression that we memorized. In example 4 I assume this is because 1/2 is very near 0, so you choose a=0, am I right? But example 4 is more general, is it because the extension of the interval you find does not change with the value of a chosen or is there another reason?
On another subject, now that I have finished the Calculus II course, can I go to the Differential Equations course or should I do  the Multivariable Calculus course first? I would appreciate your advise. Thank you.

1 answer

Last reply by: Dr. William Murray
Sat Apr 25, 2015 7:26 PM

Post by Luvivia Chang on April 21, 2015

Hello Dr. William Murray
Thanks for your patience for answering my questions in the previous lectures.
In example 5, the question asks us to use Taylor's Theorem, can the problem be done by the Alternating Series and the error be calculated by Alternating Series Error bound ?If so, which of the result will be more accurate or equally accurate?
Thanks very much.

1 answer

Last reply by: Dr. William Murray
Wed Nov 12, 2014 6:15 PM

Post by Zam Htang on November 12, 2014

what is going on this video? in calculus 2

3 answers

Last reply by: Dr. William Murray
Mon Aug 4, 2014 7:08 PM

Post by Asdf Asdf on May 17, 2014

This is the same thing as the Lagrange Error Bound right? Thanks by the way. My Calc BC exam is in a few days and I can't thank more for playing such a big role in my studying

1 answer

Last reply by: Dr. William Murray
Thu Apr 24, 2014 6:07 PM

Post by Taylor Wright on April 19, 2014

Thank you for this amazing lecture series over CalcII!!!  Do you know if there are any plans to incorporate any Engineering specific lectures in the near future such as statics, dynamics, fluids, etc.?

Thank you!

1 answer

Last reply by: Dr. William Murray
Wed Aug 28, 2013 5:51 PM

Post by William Dawson on August 25, 2013

How did you ignore the rest of the function when estimating for -x^5? What would be the problem with leaving it as a positive number, as originally written? Why wasn't it x^-5 since it was in the denominator?

3 answers

Last reply by: Dr. William Murray
Wed Aug 22, 2012 1:26 PM

Post by Mehul Patel on April 17, 2011

I got a 5 in Calc BC no thanks to!!!!!!! JK JK JK :) <<33

1 answer

Last reply by: Dr. William Murray
Thu May 30, 2013 4:21 PM

Post by Eleazar Estorga on November 7, 2010

I am having a har time understanding how to get M

Taylor Polynomial Applications

Main formulas:

Alternating Series Error Bound: If an is a series that satisfies Alternating Series Test, and we use the partial sum sn as an estimate of the total sum S (the true answer), then our error E could be positive or negative, but it’s bounded by |an+1|:

|E| ≤ |an+1|

Taylor’s Remainder Theorem: Suppose you use the Taylor polynomial Tk (x) around a to estimate f (x) at a value of x near a. Find the maximum value of f (k+1) on the interval between a and x (or an upper bound for it). Call that M. Then the

Error ≤.

Hints and tips:

  • The point of these formulas is that (among other applications) we will use Taylor polynomials to estimate the values of f (x) for some functions that would otherwise be difficult to compute. Then we will use these error formulas to determine how accurate our estimates are.

  • General principles on accuracy of Taylor Series:

    1. If they don’t converge, then they’re no good at all.

    2. But they usually converge, since 1/n! gets small quickly.

    3. They are very accurate when x is close to the center a, since (x − a)n gets small quickly. When x is far from a, (x − a)n gets big and they aren’t very accurate.

    4. This is why we use different center values of a depending on what values of x we want to use the series for.

  • When you use Taylor Series to estimate values of f (x), you need to center the series around a value of a for which f (a), f ′ (a), f ′′ (a), and so on, will be easy to compute.

  • In Taylor’s Remainder Theorem, the difficult part is often coming up with the constant M . Here are several points to remember:

    • Since M only needs to be an upper bound, not an exact value, it is OK to estimate any quantities in your expression for | f (k+1) |. However, it is important to be certain that your estimate is higher than the true value, not lower.

    • |sin x| ≤ 1, and similarly for cos x.

    • |a − b| ≤ |a| + |b|.

    • To estimate high, make the numerator as big as possible and the denominator as small as possible.

  • Sometimes we are given the error tolerance ahead of time and we must use these error formulas to reverse-engineer the number of terms we need to be accurate to within the given tolerance. (For example, an engineer building a bridge might know that she needs calculations accurate to one ten-thousandth of a meter. Or a calculator might have a display of ten decimal places, so the manufacturer must design the internal algorithms to produce answers with ten digits of accuracy.)

  • In both formulas (but especially for Taylor’s Remainder Theorem), if you are given the error tolerance, it may be hard to solve the formula explicitly for n or k. You may need to use trial and error to find the right n or k. (Remember that they are whole numbers.)

  • A common mistake is to think that the Taylor polynomial Tk (x) has k terms. k refers to the degree, not the number of terms. So, for example, the Taylor polynomial T4 (x) for f(x) = sin x centered around a = 0 is T4 (x) = xx³&frasl 6 , because the term of x4 is zero.

Taylor Polynomial Applications

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
  • Main Formulas 0:06
    • Alternating Series Error Bound
    • Taylor's Remainder Theorem
  • Lecture Example 1 3:09
  • Lecture Example 2 9:08
  • Lecture Example 3 17:35
  • Additional Example 4
  • Additional Example 5

Transcription: Taylor Polynomial Applications

OK, we want to do some more examples of Taylor Polynomial Applications.0000

The first problem here is to look at the Taylor Polynomial for cos(x) and we are cutting it off at degree 4, t4(x).0006

We want to figure out if we want that to be accurate to 2 decimal places, what values of x can you plug in there.0015

Let us start out by recalling the Taylor Polynomial for cos(x).0022

That is one that we have memorized.0028

That is 1 - x2/2! + x4/4! - x6/6! and so on.0031

We have been told to use the Taylor Polynomial t4(x).0046

That means you take the Taylor Series and you cut it off at the degree 4 term.0052

That does not mean you take 4 terms, it means you look for degree 4, there it is, and you cut if off right there.0057

So, t4(x) is 1 - x2/2! + x4/4!.0066

What we notice here is that no matter what x you plug into this series,0078

It is going to be an alternating series.0084

Even if x is positive or negative, you have even powers of x will always be positive.0087

So you have a minus, a plus, a minus, plus, and so on this series obeys the alternating series test.0097

We can use the alternating series error estimation.0106

That says that the error is less than the cut off term, the first term that you did not take.0115

So in this case, the cutoff term is x6/6!.0128

Remember, in the alternating series error bound, that was an+1.0143

An easier way to think of that is just the cutoff term, so that is x6/6!.0148

Now we have been told that we have to make this thing accurate to 2 decimal places.0155

Let us interpret that as meaning we want x6/6!, we want our maximum error to be less than 0.01.0158

We do not want an error bigger than 0.01, we want to be accurate at the second decimal place.0173

We can solve this out.0178

x6, 6! is 720, so 0.01 × 720, we got that from doing 6!.0181

So, we get abs(x6) < 7.2,0195

If we solve that for x we get x < 7.21/6.0205

That is just a value that you can throw in your calculator.0211

That turns out to be 1.3896.0216

So, the conclusion there is that the Taylor Polynomial t4(x),0222

Is an accurate estimate of cos(x) to 2 decimal places.0234

In other words, the error will be at most 0.01.0251

As long as you stick to values of x that are inside this range.0260

For values of x whose absolute value is less than 1.3896, we can plug in those values of x into the Taylor Polynomial, the 4th degree Taylor Polynomial.0268

What we will get are answers that are close to cos(x), to an error tolerance of 0.01.0295

Just to recap there, we are given a degree and we are asked to make the Taylor Polynomial accurate to 2 decimal places.0304

The first thing there is to write out the Taylor Series, cut it off so that you get the Taylor Polynomial.0315

We notice that it is an alternating series so we can use the alternating series error bound.0326

The error bound says that the error is smaller than the first term that we cut off.0330

So here that is x6/6! and we want that to be less than the allowable error, 0.01.0336

We solve that out and we get a range of x, and so those are the values of x that we can plug in and get an estimate of cos(x) accurate to 2 decimal places.0346

Our last example here is to use the Taylor polynomial for sin(x) around a=pi/3.0000

We are going to use that to estimate the sin(1 radian).0010

Then we are going to use Taylor's Remainder Theorem to estimate how accurate our approximation was.0014

The first thing to do here is to find the Taylor Polynomial for sin(x).0022

Actually that was one of the examples we did before, so I am not going to work that out again.0027

I am just going to remind you of the answer here.0030

t4(x), we did this before, earlier in the lecture.0034

was sqrt(3)/2 + 1/2 × x - pi/3.0039

This is not a Maclaurin Series, it is not centered at 0.0050

It is centered at pi/3, + sqrt(3)/4 × (x-pi/3)2 - 1/12 × (x-pi/3)3 + sqrt(3)/48 × (x-pi/3)4/0055

That was the Taylor Polynomial that we worked out on the previous example.0084

I had not done all the work again to work that out.0088

To estimate sin(1) just means you plug x=1 in there.0092

So, t4(1)=sqrt(3)/2 + 1/2 × (1-pi/3), and so on.0099

You plug in x=1 into every term here.0113

(1-pi/3)4 × sqrt(3)/48 would be your last term.0120

You plug that in and I am not going to show you the details of that,0126

That is just something that you can work out using a calculator.0130

It turns out to be, I am going to write down a lot of decimal places because it turns out that this answer is going to be pretty accurate.0133


OK, so that is our estimate for sin(1) using the Taylor Polynomial.0154

That is the first part of the problem, that was the easy part of the problem.0161

Now we have to estimate how accurate that is.0166

Remember, we are not allowed to cheat and look up sin(1) on the calculator,0168

Because the whole point of this is that we are using the Taylor Polynomial to guess what sin(1) is.0170

Instead we are going to use Taylor's Remainder Theorem, and let me remind you how that goes.0180

That says that the error is at most m/k+1! × (x-a)k+1.0188

We have to address each one of those individual terms.0208

Here the k, that is the degree of the Taylor Polynomial, so that is the 4 here. So k=4.0211

a is the place where you are centering the Taylor Series, so that is a, a=pi/3.0221

The x value that you are plugging in is 1.0231

Those are kind of the easy parts.0235

The tricky part is m, what do we do about m?0238

Well, remember, m is bound on the k+1 derivative.0242

M, we want to bound, ask how big can the k+1 derivative,0250

Well k=4, so this is the 5th derivative of f(x).0261

How big can that be on the interval between x and a.0266

So between pi/3 and 1.0280

Now, that can be a tricky problem to find out exactly what the maximum value of that derivative could be.0288

However, there is an easy way to give us a simple upper bound.0295

Note that all derivatives, including the 5th one, of f(x)=sin(x), that is the function we are talking about.0301

All of the derivatives are just, well either plus or minus sin(x), depending on where you are in the rotation.0314

Or + or - cos(x).0324

All of these derivatives are + or - sin(x) or cos(x).0327

+ or - sin(x) and + or - cos(x), they are all bounded by 1.0332

They never get bigger in absolute value than 1.0337

So, any of these derivatives, no matter what value we plug in and no matter what derivative it is, these derivatives are always less than or equal to 1.0343

This is a very common estimation when you are talking about sin(x) or cos(x), you have to make an estimate on how big the derivatives are.0360

You know that the derivative of sin(x) or cos(x) will always be either sin or cos, and sin or cos never gets bigger than 1 in absolute value.0369

So we can take m=1.0380

Let me remind you again.0382

We are trying to work out this formula, m/k+1! × x-ak+1.0385

We figured out now that we can use m=1, and k was 4, x is 1, and a is pi/3.0400

Let us work that out.0415

x-a is something we are going to have to work out.0417

x-a is 1 - pi/3.0420

That is 1 - well pi is about 3.14/3.0428

3.14/3 is definitely less than 1.05, it is safely less than that.0440

So 1 - that is safely less than 0.05.0450

That is 1/20.0461

So, I can say that this x-a term, this term of the formula is safely less than 1/20.0465

Let me plug all of these values in.0473

Our error in absolute value is < or = to, m=1, k+1! is 5!,0476

Now the x-a term, we just said that is less than 1/20. Then k+1=5.0490

So this in turn is equal to, 5! is 1/120.0500

1/205, is, 205, well 25=32 and 105 gives me 5 0's there, so we get 3,200,000 there.0512

Just to make a rough estimate, this is < or = 1 over, well 120 is less than 100.0535

Sorry, 120 is bigger than 100 so 1/120 is less than 1/100,0548

And 3,200,000 is bigger than 3,000,000 so we can safely say that 1 over that is less than 1/3,000,000.0555

So this entire thing is < or = 3,000,000 × 100.0569

So that is 1/300,000,000.0580

What that is saying is that, by Taylor's Remainder Theorem,0595

Our error is less than 1/300,000,000.0605

In other words, the value we got from plugging 1 to t4(x), t4(1),0622

Is close to the actual value of sin(1).0632

It is so close that the difference is less than 1/300,000,000.0640

What I invite you to do is to look at t4(1).0648

I will remind you that that was 0.841470985.0652

Then, actually take a calculator and work out sin(1) on your calculator.0664

I will let you plug that in on your own, and look at how close those values are to each other.0670

They really are extremely close and so we can say for sure, even without checking on our calculator, that our answer is less than 1/300,000,000.0674

Let us recap that problem.0685

What we did here was we first found the Taylor Polynomial t4(x) for f(x)=sin(x) around a=pi/3.0688

That was actually something that we did in a previous example problem.0698

You can go back and check the previous example problem to see the mechanics of that.0701

We took that Taylor Polynomial and we plugged in x=1.0705

Into that polynomial, so we did not actually work out sin(1) on a calculator.0715

We plugged in x=1 into the Taylor Polynomial, and that is where this value 0.8414 and so on came from.0720

Then to estimate the accuracy we had to look at this Taylor Remainder Formula.0729

This Taylor Remainder Formula, m/k+1! (x-a)k+1 and we filled in all of our values.0737

x is 1 because that is the value we are estimating, a=pi/3 because that is the center of the series,0745

k=4, that comes from right there, the degree of the polynomial,0752

Then m was this bound on the derivative,0760

And technically it is a bound on the 5th derivative of f.0764

But what I know is when you take derivatives of sin and cosin, they are all bound by 1.0767

So just for convenience I used m=1 here.0772

Then I worked out x-a, I said that is safely less than 1/20, so I plugged that in.0778

k+1! is 1/120, and then just to give myself round numbers, I rounded all the denominators down,0785

Which means that I am rounding the entire numbers up, which is still safe to say that the error is less than our answer of 1/300,000,000.0795

Our final conclusion there is that our estimate using the Taylor Polynomial must be close to the true value of sin(1), to within 1/300,000,000.0806

That concludes our section on applications of Taylor Polynomials, and in fact that concludes all of our Calculus 2.0820

So, this has been Will Murray, I hope you enjoyed it,0827

This is

This is and we are here to talk about applications of Taylor Polynomials.0000

The idea here is that we are going to write down some Taylor Polynomials, and plug in some values of x and we will see how close we get to the original function values.0007

That will probably make some more sense after we study some examples, but I want to give you a couple of tests that we are going to be using to check how accurate we are.0020

The first one is one that we have seen before, if you look back at the lecture on alternating series.0028

We learned the alternating series error bound.0033

What that said is that if you have a series that satisfies the conditions for the alternating series test, then if we just take a partial sum, as an estimate of the total sum, then our error can be positive or negative, but it is bounded by aN+1.0037

In other words, the first term that we do not take, the first term that we cut off.0059

So, if you are a little rusty on that, you might want to go back and look at the lecture on alternating series.0068

That is where we did some practice with error bounds for alternating series.0074

The second series we are going to use is a new one.0078

It is called Taylor's Remainder Theorem.0082

It says suppose you use the Taylor Polynomial tk(x) to estimate a function at a value of x near a.0083

Then, it gives you this formula for the error but we have to break that down piece by piece because it is a little bit complicated.0093

The first thing to look at is what is the m in this formula.0100

What you do is you find the k+1 derivative of f.0105

That is the k+1 derivative, and look at the interval between a and x, and you ask yourself, how big can that derivative get on that interval.0110

In other words, how big can that derivative get.0121

You find the maximum value, or at least an upper bound.0126

You call that upper bound m.0135

That is what the m is in this formula.0137

Then the rest of the formula looks a lot like the original Taylor Formula.0140

You have a k+1 factorial in the denominator.0148

Remember, that k comes from the degree of the Taylor Polynomial that you are looking at.0151

Then x-ak+1.0158

Again, the k is the degree of the Taylor Polynomial.0162

The a is the coming from the center of the series, so that is that a right there.0167

The x is the value that you are plugging in to use the Taylor Polynomial to guess the value of the function.0174

So that comes from that x right there.0180

That is all a little complicated, but it will probably be easier after we work out some examples.0183

Let us move straight to some examples.0188

The first one, we are going to use the Taylor Polynomial t2(x) to estimate cos(1/2).0190

The idea here is we are trying to work out cos(1/2) without a calculator or in fact...0198

If you are a programmer, if you are writing the programs for calculators, you would be using Taylor Polynomials to work these values out.0204

We are going to try to estimate the value of cos(1/2).0212

Then we are going to give an error bound to say how accurate we are.0215

We start out by remembering the Taylor Series for cos(x), which is one of those common ones we memorized.0222

cos(x) = 1 - x2/2! + x4/4! - x6/6! ...0230

So, t2(x), that means we cut off the Taylor series at the degree 2 term.0246

Because that is the degree 2 term, t2(x) says we just look at 1 - x2/2!.0257

So we cut off the series at the degree 2 term and that is the Taylor Polynomial.0270

We are going to estimate cos(1/2) using the Taylor Polynomial, that is t2(1/2).0277

That is 1 - 1/22/2, which is 1 - 1/4-2, which is 1 - 1/8. Which is 7/8. Which is 0.875.0284

That is our estimate of cos(1/2).0305

We think that cos(1/2) is approximately equal to 0.875.0309

How do we know if that is an accurate estimate.0320

Of course we could cheat and look at cos(1/2) on the calculator, but the whole point of this is to know how accurate we are without cheating.0324

Because if we know what cos(1/2) was, we would not need the Taylor Polynomial in the first place.0331

We want to give an error bound for the estimate.0337

What we are going to do is look at this series and notice that it is an alternating series, because these terms alternate from positive to negative.0340

So, we will use the alternating series error bound.0355

Which says that the error is < or = the first term that we cut off.0365

In this case that is x4/4!, and we plug in 1/2 there.0375

1/24/4!, so that is the cutoff term there, that is aN+1.0384

That in turn, that is 4!, so this is 1/24 is 1/16, 1/16 × 24.0395

Then we can just multiply that together.0412

If you want a quick off the cuff estimate of that, I know that is certainly, 16 × 24 is way bigger than 100.0415

So this is less than 1/100, considerably. 1/100 is 0.01, so our error is less than 0.01.0427

Our estimate is accurate at the very least, to 2 decimal places.0446

So, without checking on our calculator, I know that the cos(1/2) is approximately 0.875 and I know that my error there is definitely less than 0.01.0462

In fact I can say that it is even less, but I know for sure that I am certainly accurate to 2 decimal places.0480

In fact, if you check on a calculator, cos(1/2) = 0.87758.0488

So, in fact we did get the first 2 decimal places right there in our estimation there.0500

And that was without using a calculator at all, we knew that we were going to be right.0507

So, to recap there, what we did there was we took the Taylor Series for cos(x),0511

We cut it off at the degree 2 term to get the Taylor Polynomial for cos(x).0518

We plugged in 1/2, we got a value for cos(1/2) or at least an approximate value.0525

Then we noticed that the Taylor Series was an alternating series, so we could use the alternating series error bound.0530

That says you look at the cutoff term, work that out to be less than 0.01.0537

That tells us our error < 0.01.0543

Let us try another application of Taylor Polynomials.0550

Which is to do some integrals that would be very very difficult or even impossible without Taylor Polynomials.0552

Here is an integral of essentially 1/ I will write it as 1+x5.0561

That would be a really nasty integral to try to do without the use of Taylor Polynomials.0570

If you look at some of the integration techniques we learned, they would not do you very well for 1/1+x5.0575

Instead, we are going to notice that this is the same as 1/1-(-x5).0585

That is the sum of the geometric series.0593

We can write that as the sum from n=0 to infinity of (-x5)n.0596

Of course that is only for values of x within the radius of convergence.0610

The radius of convergence of that geometric series is 1.0618

So, we can write that as 1, now + -x5, so -x5.0625

+ (-x5)2, so + x10 - x15 and so on.0635

That was just the function 1/1+x5.0645

What we really want is the integral of all that.0650

The integral of 1/1+x5 dx is the integral of 1-x5 + x10 - x15 and so on dx.0655

I can integrate that just term by term so I get x - x6/6 + x11/11 - x16/16.0672

By the way, I do not need to put the arbitrary constant on here because this is a definite integral.0690

I am going to be plugging in the values of x here, so if I put an arbitrary constant on there,0697

The values of the constant would just cancel each other off.0704

What I am going to do, is evaluate this integral from 0 to 1/2 of 1/1+x5.0707

That is 1/2 - (1/2)6/6 + 1/211/11 and so on.0719

Now, I want to estimate what that series adds up to.0733

I have already been told that I want to make sure it is accurate to three decimal places.0740

I have got to make sure this thing is accurate to 3 decimal places.0748

What I notice here is that this is an alternating series, so we are going to use the alternating series error estimation.0750

We want, remember the alternating series says that the error is at most, whatever term we cut off, and we want that to be less than 0.001 because that was given to us in the problem.0765

Let us just look at these terms and see if we can figure out which one will be less than 0.001.0785

Obviously 1/2 is not less than 0.001.0795

Next possible term is 1/211/6, which is the same as 1/ 26 is 64 × 6.0799

Now remember 0.001 is 1/1000, if we ask ourselves whether that is less than 1/1000 and 64 × 6 is not bigger than 1000, so this fails.0814

Let us try the next term, 1/211/11.0834

Now, that is 1 over 211 × 11.0842

I know that 210 is about 1000, if you multiply out 210 it actually turns out to be 1024.0850

So 2 11 × 11 is certainly bigger than 1000, so this is certainly less than 1/1000.0856

So we have found the term that is less than 1/1000.0867

That means that you can use this as the cutoff term.0874

And just keep the rest of the series.0881

Let us look at the rest of the series.0887

1/2 - 1/26/6 is, if we simplify that it is 1/2 - 1/64 × 6, as we figured out before.0890

1/2 - 1/64 ×6 is 384.0908

We can write 1/2 as 192/384 - 1/384.0916

That simplifies down to 191/384.0928

So, that is our estimate for the value of this integral.0937

I know without even checking anywhere else that that answer is accurate to 3 decimal places.0942

In fact if you put 191/384 into a calculator, it turns out to be 0.497396.0950

And, if you put this integral into sophisticated integration software, you get the true answer approximates to 0.497439.0964

So, if you look at those, in fact they do match up to the first three decimal places.0982

They are not so far apart even in the next decimal place, so our answer really was accurate to within 0.001.0988

Just to recap that example, we are trying to solve this integral but it is a really nasty integral.0996

Instead what we did was we looked at the function, we managed to write it as a geometric series, then expand that into the terms of a Taylor Series.1003

Then you can integrate those terms term by term.1015

We get an answer and we plug in our values of x=1/2 and x=0.1018

Then we get what turns out to be an alternating series, and so we try to find out what term could we cut off that would be less than our desired error of 1/1000.1024

Turns out that this term is small enough, the previous term was not small enough, so this term is small enough.1038

Then we can work out the rest of the series, the stuff before that, and get our answer.1045

We know that is accurate to within 0.001.1050

Let us try another example here.1057

We want to find the Taylor Series for sqrt(x) around a=4.1060

We are going to use that to estimate the sqrt(3.8) again to within 0.001.1063

We want 3 decimal places of accuracy here.1070

I am going to find my Taylor Series, I am just going to use the generic formula for Taylor Series.1075

Make a little chart here, nth derivative of x,1079

The nth derivative, now I am going to plug in the value a=4,1084

The whole coefficient is obtained by taking that nth derivative and dividing by n!.1090

So I am going to do this for several values of n.1101

I am going to take it down to n=3 and we will see if that is enough.1105

The 0th derivative just means you take the original function, sqrt(x).1110

First derivative, if you think of that as x1/2 we will write the derivative as 1/2 x-1/2.1116

Second derivative is 1/2 × -1/2 so that is -1/4 x-3/2.1123

The third derivative is -1/4 × -3/2, that is 3/8 x-5/2.1130

Now we plug in 4 to each one of those.1144

Well sqrt(4) is just 2.1147

Here we have 4-1/2, so that is 1/sqrt(4) that is 1/2 × 1/2, so this is 1/4.1153

x-3/2, that is 43/2, that is 1/8 × -1/4 that is -1/32.1165

x-5/2, that is 45/2, sqrt(4) is 2, 25 is 32, so we get 3/8 × 32,1177

Which is 25 × 23, 28 is 256.1181

Then to get the full coefficients, we divide those numbers by the factorials.1199

So, 2/0! is just 2.1204

1/4/1! is just 1/4.1208

-1/32/2!, 2! is just 2, so that is -1/64.1211

3/256/3!, so that is divided by 6, the 3 and the 6 cancel, so you get 1/512.1220

Those are our Taylor coefficients.1245

We want to use the Taylor Remainder Theorem to estimate the error of this series.1251

Let me remind you how the Taylor Remainder Theorem works.1261

That said that the error is bounded by m/k+1! × x-ak+1.1265

There are a lot of things that we need to talk about here.1282

Let us first of all, first of all we need to figure out what k we might want to use.1286

This is just a guess at this point, so let us try, we are going to try k = 2.1292

We will see if that makes the error small enough, so we will plug k=2 into this error formula and see what we get.1302

Now, m, m comes from, remember, m comes from the maximum of that derivative.1311

We look at the k+1 derivative, so that is f the third derivative because k=2 we are going to look at the third derivative of f.1322

So the third derivative of f is, I am reading that here,1332

That is 3/8/x5/2.1338

Now to get m, you have to look at that derivative and you have to look at it on the interval between x and a.1350

Here, our x that we are interested in is 3.8. And a=4.1359

We are looking at the interval between 3.8 and 4.1367

We want to ask what is the maximum possible value of that derivative between 3.8 and 4.1375

Well, we have got an x in the denominator here.1382

So, that is going to be maximized when x is as small as possible.1388

Let me write that down.1395

This is maximized when x is as small as possible.1397

In the interval 3.8 to 4, the value you want to plug in there is the smallest possible value, so that is 3.8.1405

That is 3/8/3.85/2.1421

OK, so I can write that as 3/8, now I do not want to work out 3.85/2.1430

What I can tell you for sure though, is I am just going to give a very rough estimation.1439

I know that 3.8 is way bigger than 1.1445

I know that 3.8 in the denominator will be much smaller than 15/2, that is not 1 and 5/2, that is 15/2.1453

This is because 3.8 is way bigger than 1, so in the denominator, 3.85/2 will be less than 15/2.1467

So this is equal to, just simplifies down to 3/8.1477

We are going to take m=3/8.1486

Our error using this formula now, is less than or equal to 3/8, that is from m.1493

k+1!, remember we said we were going to try k=2.1508

So 2+1! is 3!, that is 3!. x-1, these are the values of x and a.1514

So 3.8-4k+1, so that is 3, because k=2.1525

This simplifies down a bit to 3/8 × 6, because 3! is 6.1537

Now 3.8-4 is -0.2, and the absolute value of that is just 0.23.1547

0.23 is 0.008, over, the 3 and the 6 cancel, so we get 16.1560

This gives us 0.008/16, which is 0.0005.1572

The important thing here is 0.0005 is less than 0.001.1582

OK, so let us keep going with this on the next page.1592

What we did was we tried k=2.1598

Then we looked at the error and we did a bunch of simplifications,1604

But what we got was that it was less than 0.001.1611

That worked OK, that was acceptable for the error tolerance that we were given.1615

If that had not worked, I kind of pulled out k=2 seemingly randomly and that was actually because I worked out this problem ahead of time, of course.1620

If that had not worked, if it does not work, You just try a bigger value of k.1632

You just try k=3 or 4, just keep trying values of k until you get a value that works.1645

In this case k=2 works.1653

That means we have to find the Taylor Polynomial of degree 2.1655

What I am going to do is remind you of those coefficients from the table on the previous page.1661

That was 2, this was the cn, 2 and then 1/4, and then -1/64.1667

1/5/12, that was the last part of the table from the previous page.1676

t2(x) means we read off these coefficients and we attach each one to a power of x.1681

So, 2 × x0 is just 1.1690

+ 1/4 ×, sorry not a power of x, but x-a, so that is x-41.1694

We go up to the 2 term, so that is -1/64 × x-42, and then we cut if off there, because we are using k=2.1704

Finally we want to estimate the sqrt(3.8), so we are going to plug 3.8 into t2(x).1719

3.8 gives us 2+1/4, 3.8 - 4 = -0.2.1729

-1/64 × -0.22.1743

We can simplify that a little bit, this is 2 - 0.2/4, 0.22 is, sorry 0.04/64,1752

At this point it is probably worthwhile to just put these numbers into a calculator. We get the estimate of 1.94937.1772

That is our estimate for f(3/8), in other words sqrt(3.8).1785

That was a lot of work for one problem, let me recap what we did there.1795

First of all we wrote down the terms for the Taylor Series.1799

Then we had to use Taylor's Remainder Formula, the error formula, to try to find where should we cut this series off, so that our error is less than 0.001.1804

We kind of arbitrarily tried k=2, it turned out to work, but if it did not work we would try a different value of k,1820

We worked out Taylor's Error formula.1828

That involved finding the m term, the maximum value of the derivative, plugging in all the other terms to the error formula, and we get that it was less than 0.001.1830

That tells us that we can use the Taylor Polynomial of degree 2, so we fill that in using these coefficients that we worked out.1842

Then we used that Taylor Polynomial and we plug in our value of 3.8, and we work it down and find our answer for our estimate of 3.8.1852

We will try some more examples of that later.1863