For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

## Discussion

## Study Guides

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## Table of Contents

## Transcription

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### Taylor Polynomial Applications

**Main formulas:**

**Alternating Series Error Bound:** If ∑*a _{n}* is a series that satisfies Alternating Series Test, and we use the partial sum

*s*as an estimate of the total sum

_{n}*S*(the true answer), then our error

*E*could be positive or negative, but it’s bounded by |

*a*

_{n+1}|:

|*E*| ≤ |*a*_{n+1}|

**Taylor’s Remainder Theorem:** Suppose you use the Taylor polynomial *T _{k}* (

*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:

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

But they usually converge, since 1/

*n*! gets small quickly.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.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

*T*(_{k}*x*) has*k*terms.*k*refers to the degree, not the number of terms. So, for example, the Taylor polynomial*T*_{4}(*x*) for*f*(*x*) = sin*x*centered around*a*= 0 is*T*_{4}(*x*) =*x*−*x*³&frasl 6 , because the term of*x*^{4}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

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### 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, t _{4}(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 - x ^{2}/2! + x^{4}/4! - x^{6}/6! and so on.*0031

*We have been told to use the Taylor Polynomial t _{4}(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, t _{4}(x) is 1 - x^{2}/2! + x^{4}/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 x ^{6}/6!.*0128

*Remember, in the alternating series error bound, that was a _{n}+1.*0143

*An easier way to think of that is just the cutoff term, so that is x ^{6}/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 x ^{6}/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

*x ^{6}, 6! is 720, so 0.01 × 720, we got that from doing 6!.*0181

*So, we get abs(x ^{6}) < 7.2,*0195

*If we solve that for x we get x < 7.2 ^{1/6}.*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 t _{4}(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 4 ^{th} 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 x ^{6}/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, t _{4}(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

*0.841470985.*0142

*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-a ^{k+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

*1/20 ^{5}, is, 20^{5}, well 2^{5}=32 and 10^{5} 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 t _{4}(x), t_{4}(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 t _{4}(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 t _{4}(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 5 ^{th} 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 educator.com.*0829

*This is educator.com 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 t _{k}(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-a ^{k+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 t _{2}(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 - x ^{2}/2! + x^{4}/4! - x^{6}/6! ...*0230

*So, t _{2}(x), that means we cut off the Taylor series at the degree 2 term.*0246

*Because that is the degree 2 term, t _{2}(x) says we just look at 1 - x^{2}/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 t _{2}(1/2).*0277

*That is 1 - 1/2 ^{2}/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 x _{4}/4!, and we plug in 1/2 there.*0375

*1/2 _{4}/4!, so that is the cutoff term there, that is aN+1.*0384

*That in turn, that is 4!, so this is 1/2 _{4} 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+x ^{5}.*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+x ^{5}.*0575

*Instead, we are going to notice that this is the same as 1/1-(-x ^{5}).*0585

*That is the sum of the geometric series.*0593

*We can write that as the sum from n=0 to infinity of (-x ^{5})^{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 + -x ^{5}, so -x^{5}.*0625

*+ (-x ^{5})^{2}, so + x^{10} - x^{15} and so on.*0635

*That was just the function 1/1+x ^{5}.*0645

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

*The integral of 1/1+x ^{5} dx is the integral of 1-x^{5} + x^{10} - x^{15} and so on dx.*0655

*I can integrate that just term by term so I get x - x ^{6}/6 + x^{11}/11 - x^{16}/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+x ^{5}.*0707

*That is 1/2 - (1/2) ^{6}/6 + 1/2^{11}/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 1/2 ^{11}/6, which is the same as 1/ 2^{6} 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, 1/2 ^{11}/11.*0834

*Now, that is 1 over 2 ^{11} × 11.*0842

*I know that 2 ^{10} is about 1000, if you multiply out 2^{10} 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 - 1/2 ^{6}/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, n ^{th} derivative of x,*1079

*The n ^{th} derivative, now I am going to plug in the value a=4,*1084

*The whole coefficient is obtained by taking that n ^{th} 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 0 ^{th} derivative just means you take the original function, sqrt(x).*1110

*First derivative, if you think of that as x ^{1/2} 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 4^{3/2}, that is 1/8 × -1/4 that is -1/32.*1165

*x ^{-5/2}, that is 4^{5/2}, sqrt(4) is 2, 2^{5} is 32, so we get 3/8 × 32,*1177

*Which is 2 ^{5} × 2^{3}, 2^{8} 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-a ^{k+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/x ^{5/2}.*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 3/8/3.8 ^{5/2}.*1421

*OK, so I can write that as 3/8, now I do not want to work out 3.8 ^{5/2}.*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 1 ^{5/2}, that is not 1 and 5/2, that is 1^{5/2}.*1453

*This is because 3.8 is way bigger than 1, so in the denominator, 3.8 ^{5/2} will be less than 1^{5/2}.*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 3.8-4 ^{k+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 0.2 ^{3}.*1547

*0.2 ^{3} 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 c _{n}, 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

*t _{2}(x) means we read off these coefficients and we attach each one to a power of x.*1681

*So, 2 × x ^{0} is just 1.*1690

*+ 1/4 ×, sorry not a power of x, but x-a, so that is x-4 ^{1}.*1694

*We go up to the 2 term, so that is -1/64 × x-4 ^{2}, 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 × -0.2 ^{2}.*1743

*We can simplify that a little bit, this is 2 - 0.2/4, 0.2 ^{2} 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

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 educator.com 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 educator.com!!!!!!! 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