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

### Taylor Series and Maclaurin Series

**Main definitions:**

**Definitions:** The __Taylor
Series__ for a function *f* (*x*) around a center value *a*
is the power series

.

Here *f* ^{(n)}
(*a*) represents the *n*-th derivative of *f*, with *a*
plugged in.

The __Maclaurin Series__ for *f*
(*x*) is just the special case of the Taylor Series around the
center value *a* = 0:

.

The __Taylor polynomial__ is what
you get when you cut off the Taylor Series at the degree *k*
term:

=

**Hints and tips:**

In many cases, you do not want to use the formulas above to find the Taylor Series of a function, because the derivatives get too messy. Instead, start with some known Taylor Series for some common function and derive other series from the known series using the following techniques:

Algebraic manipulations, e.g. multiplying by

*x*.Substitutions, e.g. replacing

*x*by 2*x*or*x*² .Derivatives and integrals.

Multiplying or dividing two series together.

You should memorize the Maclaurin Series for

*e*^{x}, sin*x*, and cos*x*at the very least, and probably for 1/(1−*x*), arctan*x*, and ln(1 −*x*) as well.Sometimes you cannot find the general pattern for a Taylor Series, especially those that are not centered at

*a*= 0. However, you can still find the first few terms, and this might be enough for computations.The Taylor series for a polynomial is just the polynomial itself. 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⁄ 6 , because the term of*x*^{4}is zero.

### Taylor Series and Maclaurin Series

- Find derivatives and f
^{n}(0) -
f(x) = [1/(2 − x)] f(0) = [1/2] f′(x) = [1/((x − 2) ^{2})]f′(0) = [1/4] f"(x) = [2/((x − 2) ^{3})]f"(0) = [1/8] f"′(x) = [6/((x − 2) ^{4})]f"′(0) = 116

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n})∑

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n}) = [1/2] + [x/4] + [(x

^{2})/8] + [(x

^{4})/32] + ...

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n})[1/(2 − x)] = [1/2] + [x/4] + [(x

^{2})/8] + [(x

^{4})/32] + ... = ∑

_{n = 0}

^{∞}(x − 1)

^{n}

- Find derivatives and f
^{n}(0) -
f(x) = [1/(x + 1)] f(0) = 1 f′(x) = − [1/(( x + 1 ) ^{2})]f′(0) = − 1 f"(x) = [2/(( x + 1 ) ^{3})]f"(0) = 2 f"′(x) = − [6/(( x + 1 ) ^{4})]f"′(0) = − 6

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n})∑

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n}) = 1 − [x/1] + [(2x

^{2})/2!] − [(6x

^{3})/3!] + ... = 1 − x + x

^{2}− x

^{3}+ ...

- Find General Form of ∑
_{n = 0}^{∞}[(f^{n}(0))/n!] ( x^{n})

^{2}− x

^{3}+ ... = ∑

_{n = 0}

^{∞}( − 1)

^{n}x

^{n}

^{ − x}

- Find derivatives and f
^{n}(0) -
f(x) = e ^{x}f(0) = 1 f′(x) = − e ^{x}f′(0) = 1 f"(x) = e ^{x}f"(0) = 1 f"′(x) = − e ^{x}f"′(0) = 1

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n})∑

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n}) = 1 − [x/1] + [(x

^{2})/2!] − [(x

^{3})/3!] + ... = 1 − x + [(x

^{2})/2] − [(x

^{3})/6] + ...

^{ − x}?

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n})e

^{ − x}= 1 − x + [(x

^{2})/2] − [(x

^{3})/6] + ... = ∑

_{n = 0}

^{∞}[(( − 1)

^{n}x

^{n})/n!]

^{x})]

- Find derivatives and f
^{n}(0) -
f(x) = [x/(e ^{x})]f(0) = 0 f′(x) = − [(x − 1)/(e ^{x})]f′(0) = 1 f"(x) = [(x − 2)/(e ^{x})]f"(0) = − 2 f"′(x) = − [(x − 3)/(e ^{x})]f"′(0) = 3

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n})∑

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n}) = 0 + [x/1!] − [(2x

^{2})/2!] + [(3x

^{3})/6!] + ... = x − x

^{2}+ [(x

^{3})/2] − [(x

^{4})/6] + ...

^{x})]?

- Find General Form of ∑
_{n = 0}^{∞}[(f^{n}(0))/n!] ( x^{n})

^{x})] = x − x

^{2}+ [(x

^{3})/2] − [(x

^{4})/6] + ... = x∑

_{n = 0}

^{∞}[(( − 1)

^{n}x

^{n})/n!]

- Find derivatives and f
^{n}(0) -
f(x) = ln|x + 1| f(0) = 0 f′(x) = [1/(x + 1)] f′(0) = 1 f"(x) = − [1/(( x + 1 ) ^{2})]f"(0) = 1 f"′(x) = [2/(( x + 1 ) ^{3})]f"′(0) = [1/4]

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n})∑

_{n = 0}

^{∞}[(f

^{n}(0))/n!] ( x

^{n}) = 0 + [x/1] − [(x

^{2})/2!] + [(2x

^{3})/3!] − [(6x

^{4})/4!] + ... = x − [(x

^{2})/2] + [(x

^{3})/3] − [(x

^{4})/4] + ...

- Find General Form of ∑
_{n = 0}^{∞}[(f^{n}(0))/n!] ( x^{n})

^{2})/2] + [(x

^{3})/3] − [(x

^{4})/4] + ... = ∑

_{n = 0}

^{∞}[(( − 1)

^{n}x

^{n})/n]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Taylor Series and Maclaurin Series

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.

### College Calculus 2 Online Course

### Transcription: Taylor Series and Maclaurin Series

*Here we were asked to find the example, to find the McLauren series for e ^{x}.*0000

*This is one where we do not have anything memorized yet.*0009

*So, we are going to work out n and the nth derivative of x.*0014

*Since it is a McLauren series, we are going to plug in f _{n}(0).*0017

*Then finally our coefficient is the nth derivative of 0 divided by n!.*0021

*We will work this out for a few values of n.*0032

*f _{n}(x), well e^{x}, that is very easy, that is e^{x} all the way down.*0038

*So, f _{n}(0), that is just plugging 0 into e^{x}, so that is 1 all the way down.*0043

*cN is the coefficient, you divide that by n!.*0054

*We get 1/1, because 0 and 1! are both 1, 1/2!, 1/3!, and so on.*0060

*The McLauren series is, remember, we take the coefficient f _{n}(a)/n!/x-a^{n}.*0074

*Then for the McLauren series the a is just 0.*0092

*So that means each one of these coefficients gets multiplied by a power of x.*0094

*So, it is 1 + 1 × x + 1/2! × x ^{2} × 1/3! x^{3}, and so on.*0099

*So, we get the McLauren series for e ^{x} is the sum from 0 to infinity of just x^{n}/n!.*0115

*That, again, is such a common function and such a common Taylor Series.*0130

*It comes up in so many different places and in particular Calculus classes, that that is one you should really memorize.*0135

*The first time you ever work it out, you have to use this derivative formula.*0146

*You have to write down the derivatives, plug in x=0 and then you get these coefficients,*0152

*And you multiply each coefficient by a power of x.*0158

*That is kind of the first time that you work it out.*0162

*From the on it is probably worth memorizing that the McLauren series for e ^{x}, is just x^{x}/n!.*0164

*OK, for this example we are asked to find the McLauren series for f(x) = (e ^{x})^{2}.*0000

*You could try to write down the derivative, so let us try to write down a couple of derivatives.*0007

*We know f(x) = e ^{x}. We are going to write down n fn(x). Oops, f(x) = (e^{x})^{2}.*0012

*So f0(x) is just (e ^{x})^{2}.*0031

*First derivative is just (e ^{x})^{2} × 2x.*0036

*The second derivative is 2x × derivative of (e ^{x})^{2}, so that is 2x × 2x × (e^{x})^{2}.*0041

*Plus (e ^{x})^{2} × derivative of 2x that is 2 (e^{x})^{2}.*0053

*That can simplify down to 4x ^{2} (e^{x})^{2},*0062

*I guess 4x ^{2} + 2, we could combine those terms (e^{x})^{2}.*0068

*What you can see is that this is already getting messy and it is going to be difficult to find a pattern.*0076

*Taking the third derivative, things are getting more and more complicated.*0082

*We are already getting into something messy and complicated and in particular, it is going to be difficult to find a pattern.*0088

*What I want to show you is a better way to find this McLauren's series.*0098

*A better way than trying to write down all of these derivatives, which are getting messy and complicated.*0104

*A better way is to remember that we just figured out a McLauren series for e ^{x}.*0114

*e ^{x} was the sum from 0 to infinity of x^{n}/n!.*0121

*A much easier way than writing down all of these derivatives is to substitute in x ^{2} for x.*0136

*So (e ^{x})^{2} is the sum form n=0 to infinity of x^{2}^{n}/n!.*0145

*You could write that as the sum from n=0 to infinity of x ^{2n}/n!.*0158

*If you want, you could expand this out term by term if you plug in n=0, you get 1.*0168

*n=1 you get +x ^{2}/1.*0176

*n=2 gives you x ^{4}/2!.*0182

*n=3 gives you x ^{6}/3!, and so on.*0187

*Notice that that is just what you would have gotten if you had looked at the series for e ^{x}.*0193

*1+x+x ^{2}/2!+x^{3}/3!.*0199

*If you had taken that series and just changed x into x ^{2}, that would have given you the series we had just gotten by expanding it out.*0208

*That immediately gives us a very McLauren series for (e ^{x})^{2}.*0220

*Much easier than writing down all of these derivatives which get messy and very hard to write a pattern.*0227

*The moral there that a lot of calculus students sometimes miss is that,*0236

*Using this derivative process of finding the McLauren series or the Taylor Series is often the slowest and most difficult way, to find the McLauren series.*0240

*A much better way is to have some prefabricated series at your disposal.*0252

*By that I mean, you have memorized some series, some key series to start with.*0258

*Then when you want to find other series, you just work them out based on the series you already have memorized.*0262

*In this case, you already had memorized a series for e ^{x}, and when you have to find (e^{x})^{2} you just do a quick substitution, and we get our series for (e^{x})^{2}*0271

*Hi, this is educator.com and we are here to look at Taylor Series and Maclaurin Series.*0000

*So, a couple definitions to get us started here.*0008

*You start with a function f(x), and a value a, the center value.*0012

*Then you form this power series that we call the Taylor series of x.*0018

*You take the nth derivative, so that is the n ^{th} derivative there.*0023

*Of the function, you plug in a, you divide it by n!, and then you multiply by x-a ^{n}.*0030

*You kind of want to think about this entire term here as being a coefficient.*0041

*We will call that c _{n}, that is the n^{th} coefficient of the Taylor Series.*0047

*Then the x-a ^{n} part is the power part of the Taylor series.*0052

*You will also hear about McLauren series, that is the exact same thing as the Taylor Series.*0060

*That just means that you take the special case where a=0.*0068

*So McLauren Series is just a special case of the Taylor Series.*0070

*You just plug in a=0, and so you get something slightly simpler there.*0073

*We are also going to talk about Taylor Polynomials.*0080

*The formula for the Taylor Polynomial looks exactly the same for the Taylor series,*0087

*Except what you do is instead of running it to infinity, you cut the thing off at the degree k term.*0092

*So, you call this t _{k}(x), and what that means is, essentially that you get this thing that only runs up to x^{k} power.*0101

*You have these coefficients that give you a polynomial of degree.*0117

*Well, usually it will be degree K, but if that last term, if the coefficient happens to be 0, it could be lower degrees.*0130

*So I will see degree less than or equal to k.*0138

*So the Taylor Polynomial is a polynomial of degree k, but what it represents is you just take the first few terms of the Taylor Series,*0140

*And you run it up until you see an x ^{k} term.*0149

*Let us try some examples there.*0153

*The first example is to find the Maclaurin series for f(x) = cos(x).*0158

*That means, remember, the formula there is the n ^{th} derivative of f at a.*0163

*Well a=0 here, because it is a Maclaurin Series.*0171

*fN(0)/n! × normally x-a ^{n}, but since a=0 it is just x^{n}.*0174

*I want to figure out what these coefficients are, the c _{n}'s.*0184

*I will make a little chart here.*0188

*n and the n ^{th} derivative of x, then I will plug in the n^{th} derivative of 0.*0191

*Then I will figure out what the coefficient n is.*0198

*That is the same as f _{n}(0) divided by n factorial.*0200

*When n=0, we have the 0 derivative of f, which just means the original function.*0208

*The 0 derivative is just cos(x).*0215

*cos(0)=1, and 1/0!, remember 0! is just 1. So 1/0!=1.*0218

*When n=1, we take the first derivative, that is -sin(x).*0231

*If you plug in 0 to that, it is 0, so the coefficient is just 0.*0236

*When n=2, derivative of -sin is -cos(x).*0241

*Plug in 0 you get -1. So -1/2! is -1/2.*0247

*When n=3, we get the derivative of -cos is just sin(x).*0255

*Plug in 0 to that and you get 0. 0/3!=0.*0264

*When n=4, we get cos(x) and so that derivative when you plug in 0 is 1.*0270

*1/4! is 1/4!, if you want you can write that as 1/24.*0279

*cos(x) is, we take these coefficients, and you attach on the relevant powers of x.*0290

*That is 1 × x ^{0} + 0 × x^{1}.*0302

*So I will leave that out, - 1/2, 1/2! × x ^{2}, + 0 x^{3}, + 1/4! × x^{4}.*0307

*Then you can start to see the pattern here, of course the derivatives of sin and cos are going to repeat themselves.*0325

*The next term will be 0 x ^{5} - 1/6! × x^{6},*0333

*0 x ^{7} + and so on.*0341

*So, as a series there, we can write this as the sum from 0 to infinity.*0345

*I just want to catch the even powers, because all of the odd ones are 0.*0352

*So I can write x ^{2n}/2n!.*0358

*I am going to write -1 ^{n} because that makes it alternative positive and negative there.*0366

*That is our Maclaurin series for cos(x).*0374

*Now, there are several common Maclaurin series that it is probably worth memorizing for at least as long as you are a Calculus student.*0384

*This is one of them, cos(x) is one you should really memorize.*0395

*You should not have to work out this whole chart every time.*0397

*You should probably know the answer because it will almost certainly come up in your Calculus class.*0403

*Let us try another one. f(x) = sin(x).*0409

*Again we want to find the McLauren series.*0413

*One way to do that is to write a whole chart, like we did in the previous example, *0417

*Where you write the derivatives, you plug in 0 every time, then you divide by the factorials, you string them together and you make a series.*0421

*That is a lot of work; instead, we are going to find something easier,*0430

*Which is to notice that sin(x)=derivative of cos(x).*0435

*Except for one negative sign that I have to include there.*0446

*The point there is that I remember a series for cos(x) because we just did it in the last problem.*0452

*We memorized that series. Let me write down the series that I just memorized for cos(x).*0461

*1 - x ^{2}/2! + x^{4}/4! - x^{6}/6! and so on.*0468

*So we get -, now I am just going to take the derivative of that.*0482

*Derivative of 1 is just 0.*0485

*Derivative of x ^{2}/2! is just x. Sorry 2x/2!*0488

*+ now the derivative of x ^{4} + 4x^{3}/4!.*0503

*- derivative of x ^{6} is - x^{5}/6! and so on.*0511

*So you want to go ahead and distribute the - sign so we get +, now 2x/2!,*0520

*2! is just 2 so this is just x-, because of the - sign on the outside, 4x ^{3}/4!.*0529

*Well, 4/4!, 4! is 4 × 3 × 2 × 1.*0540

*That 4 in the numerator just cancels away that first 4 and leaves us with 3!.*0548

*Now we have got x ^{5}, 6 in the numerator and 6! in the denominator.*0555

*Well the 6 in the numerator just cancels the 6 out of the factorial, and we get 5!*0562

*You can see the pattern here.*0569

*The next one is going to be, sorry this x ^{5}/5! should have been positive,*0570

*Because it was negative before but we had a negative on the outside so those two negatives cancel. *0577

*You can see that the next term following the same pattern is going to be,*0581

*x ^{7}/7!, and so on.*0586

*If you want to write that in a nice closed summation form.*0591

*We have to find a way of capturing the odd numbers.*0598

*But, that is x ^{2n+1}/(2n+1)!.*0601

*And, -1 ^{n} will make it alternate positive and negative.*0613

*That is our Maclaurin series for sin(x).*0621

*Again, this is such a common one.*0626

*Sin(x) is such a common function that this one is worth memorizing as well.*0628

*The point of asking you to memorize these things is that later on, we could get more difficult Taylor Series, and Maclaurin series.*0636

*It helps if you kind of have a stock of pre-fabricated examples and then you can use those examples to build up more complicated ones*0645

*By taking derivatives, making substitutions, doing other algebraic tricks,*0656

*And you will not have to work everything out from first principles,*0662

*Taking lots of derivatives and so on.*0667

*In particular, in this case, we figured out sin(x) by using the fact that we had already worked out the cos(x) before.*0669

*This is really exploiting the work we did before.*0680

*We remember the answer to that.*0683

*Then we can just take its derivative and figure out a Taylor Series for sin(x).*0688

*So, example 3 is to find the Taylor Polynomial t _{4}(x) for sin(x).*0698

*This time we are not centering it around 0, we are centering it around a = pi/3.*0705

*It is not a McLaurin Series, it is a full-blown Taylor Series.*0711

*This time I am going to start with my chart.*0716

*n f _{n}(x), the n^{th} derivative, f_{n}(pi/3).*0718

*Then the full coefficient c _{n}, which is f_{n}(pi/3)/n!.*0730

*The reason I am setting that up is I am remembering the master formula for Taylor Series, which remember is the n ^{th} derivative x, of a, you plug in x=a,*0742

*Divided by n! multiplied by the power part x-a ^{n}.*0755

*That is the master formula for Taylor Series and to get the Taylor Polynomial you just cut that off at the k term.*0760

*Let us work out what these values are for the coefficients, for the first few values of n.*0774

*I will work this out for n=0, 1, 2, 3, & 4.*0785

*That is because we only have to go up to the k=4 term, or n=4.*0789

*So sin(x), my derivatives of sin(x), well the 0 derivative is just sin(x) itself.*0796

*You have not taken any derivatives. *0803

*First derivative is cos(x). Second derivative is derivative of cosine, which is -sin(x). Derivative of that is -cos(x).*0805

*Then it starts to repeat at sin(x), if you plug in pi/3 to each of these, the sin(pi/3) is sqrt(3/2).*0817

*Cos(pi/3) = 1/2.*0830

*-sin = -sqrt(3/2),*0833

*-cos is -1/2.*0836

*Then sine is sqrt(3/2) again.*0840

*Now the coefficients c _{n} says you take those derivative values and you divide them by n!. Well 0! is just 1.*0844

*So, we get sqrt(3/2), we are not dividing by anything except 1, and 1! is just 1 so we get 1/2 here.*0855

*But now 2! is 2. We get -sqrt(3/2)/2!, so that is -sqrt(3)/4.*0864

*3!=6, so we get -1/2/6, that is -1/12.*0877

*4! is 24, so we get sqrt(3)/2/24, so sqrt(3)/48.*0884

*I am going to take those coefficients and plug them into my formula for the Taylor Polynomial.*0898

*I will have an n=0 term, an n=1 term, n=2, n=3, and n=4 term.*0906

*The n=0 says that you take sqrt(3)/2 and I am getting that from here.*0917

*Then you multiply by x-a ^{0} which is just 0.*0924

*n=1 says we take the 1/2 × x-a is pi/3, so pi/3 ^{1}.*0931

*n=2 gives us -sqrt(3)/2 × x - pi/3. To the 2 power. That is an x-pi/3.*0942

*n=3 gives us -1/12.*0962

*x-pi/3 _{3} + sqrt(3)/48 × x-pi/3^{4}.*0967

*That, all of that together, is t4(x). All of that together is our solution.*0985

*To recap, what we did was we invoked this formula for the Taylor Polynomials, which again is just the formula for Taylor Series.*0998

*We cut it off. Instead of going to infinity, we cut it off at the k term, because this is a=pi/3, it is not a Maclaurin series.*1010

*It is not something where we can use something that we have memorized.*1019

*We just kind of go through and work out these coefficients one by one using the derivative formula.*1022

*Then we plug everything back into the formula where we attach the powers of x-a ^{n},*1028

*And we get our Taylor Polynomial.*1035

*We will see how we can use these later on in the next lecture on applications of Taylor Polynomials.*1037

*We will try some more examples later on.*1045

*OK, we are here working on examples of Taylor and Maclaurin series.*1050

*The example we are given here is the Maclaurin Series for f(x) = sec(x).*1056

*The way you want to think about that is you could remember the generic formula for Taylor Series,*1065

*Where you take the n ^{th} derivative of f at x over n! × x-a^{n}.*1074

*That is the generic formula for Taylor Series. *1084

*The problem with that is if you start taking lots of derivative of sec(x) it is going to be really messy because those derivatives do not repeat.*1092

*They get uglier and uglier.*1095

*All we are asked to do here is find 3 non-zero terms, so what I am going to do is exploit the fact that sec(x) is 1/cos(x).*1097

*I remember a Maclauran series for cos(x) because we have worked it out earlier and I memorized the answer.*1114

*That is 1-x ^{2}/2+x^{4}/4! is 24.*1125

*So, I am filling that in because we already worked out the series for cos(x) and we memorized it.*1140

*It is good if you memorize the series for some of these common ones like sin(x) and cos(x) and e ^{x} because they come up so often.*1149

*This is an example of exploiting one we have used before. *1157

*Our sec(x) is 1/c0s(x).*1160

*We are going to do a little long division of polynomials just like you did in high school.*1162

*We do 1-x ^{2}/2 + x^{4}/24. That gets divided into 1.*1168

*I am just going to fill in, think of 1 as a big polynomial,*1182

*So I am going to fill in, 1 + 0x ^{2} + 0x^{4}, I am only filling in even terms,*1185

*Because I only have even terms in the cos(x), so we are not going to need to use any odd terms here.*1194

*Now we just do long division of polynomials which you might have learned in high school but you might be a little bit rusty. *1200

*We look at these terms, 1 and 1, so we get 1 there. *1210

*Now we multiply 1 by the whole thing, so 1 - x ^{2}/2 + x^{4}/24,*1214

*And then we subtract just like we do long division of numbers,*1225

*So that is + x ^{2}/2, then we multiply across there, x^{2}/2.*1228

*x ^{2}/2 × -x^{2}/2 is -x^{4}/4 + x^{6}/48.*1237

*We subtract again and the x ^{2}/2's cancel.*1258

*x ^{4}/4, I can write that as 6/24.*1263

*So, we have 6/24 - 1/24, that gives us 5/24 × x ^{4} - x^{6}/48*1269

*I was only asked to find three non-zero terms,*1285

*So I just need to find one more.*1289

*That would be then, if we look at the leading term 5/24 x ^{4}.*1292

*So, that is our answer.*1302

*sec(x) = 1+x ^{2}/2 + 5x^{4}/24.*1305

*That is really just the first few terms of the Taylor Series.*1320

*To indicate that there is more to come there, I will put a ...*1323

*The question just asked us to find the first 3 non-zero terms. That is what they are.*1328

*This is an example of a problem where it would be quite difficult to find a general pattern, and to write it out in sigma form,*1336

*That is probably why the question only asked us to find the first few terms.*1345

*So, we just found the first few terms by manipulating these polynomials, using the fact that we already knew what the Taylor Series for cos(x) was.*1349

*Then we manipulated the polynomials, and we got the first few terms of a Taylor Series for sec(x).*1361

3 answers

Last reply by: Dr. William Murray

Sat Apr 25, 2015 7:34 PM

Post by Luvivia Chang on April 18 at 09:22:01 AM

Hello Dr William Murray

Before this lecture, I regarded Cn as a constant in the series. But judging from what I learned from this lecture, it seems that Cn changes as n changes. Is that right?

And I find it tricky to use x in the series. Because x has long been used as an unknown or variable in many algebra problems such as functions and equations. But it seems that x here is a coefficient and that only if x has a certain value will the whole series be determined. Am I understanding it correctly?

Thank you.

1 answer

Last reply by: Dr. William Murray

Tue May 28, 2013 7:06 PM

Post by Luis Chanez on May 27, 2013

You are very smart professor Murray. I past my calculus II class

I am looking forward in taking differential equations with you

Thanks fort everything :)

1 answer

Last reply by: Dr. William Murray

Mon Dec 3, 2012 5:29 PM

Post by Louise Barrea on December 3, 2012

Isn't there a mistake a the end of the lecture? When sec x = 1 + .... instead of sec x = 1/ (1+ ....).

1 answer

Last reply by: Dr. William Murray

Mon Dec 3, 2012 5:27 PM

Post by khadar mire on November 14, 2011

Many thanks

2 answers

Last reply by: Dr. William Murray

Tue Nov 20, 2012 6:41 PM

Post by nicholas devos on April 18, 2011

there is a problem at 15:48. Where n=2, the denominator should be 4, not 2.