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### 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 2x or x² .

• Derivatives and integrals.

• Multiplying or dividing two series together.

• You should memorize the Maclaurin Series for ex , 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 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⁄ 6 , because the term of x4 is zero.

### Taylor Series and Maclaurin Series

Find the MacLaurin Series for f(x) = [1/(2 − x)]
• Find derivatives and fn(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
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = [1/2] + [x/4] + [(x2)/8] + [(x4)/32] + ...
What is the MacLaurin Series representaiton for [1/(2 − x)]?
Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )[1/(2 − x)] = [1/2] + [x/4] + [(x2)/8] + [(x4)/32] + ... = ∑n = 0 (x − 1)n
Find the MacLaurin Series for f(x) = [1/(x + 1)]
• Find derivatives and fn(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
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 1 − [x/1] + [(2x2)/2!] − [(6x3)/3!] + ... = 1 − x + x2 − x3 + ...
What is the MacLaurin Series representaiton for [1/(x + 1)]?
• Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )
[1/(x + 1)] = 1 − x + x2 − x3 + ... = ∑n = 0 ( − 1)nxn
Find the MacLaurin Series for f(x) = e − x
• Find derivatives and fn(0)
•  f(x) = ex
 f(0) = 1
 f′(x) = − ex
 f′(0) = 1
 f"(x) = ex
 f"(0) = 1
 f"′(x) = − ex
 f"′(0) = 1
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 1 − [x/1] + [(x2)/2!] − [(x3)/3!] + ... = 1 − x + [(x2)/2] − [(x3)/6] + ...
What is the MacLaurin Series representaiton for e − x?
Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )e − x = 1 − x + [(x2)/2] − [(x3)/6] + ... = ∑n = 0 [(( − 1)nxn)/n!]
Find the MacLaurin Series for f(x) = [x/(ex)]
• Find derivatives and fn(0)
•  f(x) = [x/(ex)]
 f(0) = 0
 f′(x) = − [(x − 1)/(ex)]
 f′(0) = 1
 f"(x) = [(x − 2)/(ex)]
 f"(0) = − 2
 f"′(x) = − [(x − 3)/(ex)]
 f"′(0) = 3
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 0 + [x/1!] − [(2x2)/2!] + [(3x3)/6!] + ... = x − x2 + [(x3)/2] − [(x4)/6] + ...
What is the MacLaurin Series representaiton for [x/(ex)]?
• Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )
[x/(ex)] = x − x2 + [(x3)/2] − [(x4)/6] + ... = x∑n = 0 [(( − 1)nxn)/n!]
Find the MacLaurin Series for f(x) = ln(x + 1)
• Find derivatives and fn(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]
Apply ∑n = 0 [(fn(0))/n!] ( xn )∑n = 0 [(fn(0))/n!] ( xn ) = 0 + [x/1] − [(x2)/2!] + [(2x3)/3!] − [(6x4)/4!] + ... = x − [(x2)/2] + [(x3)/3] − [(x4)/4] + ...
What is the MacLaurin Series representaiton for ln(x + 1)?
• Find General Form of ∑n = 0 [(fn(0))/n!] ( xn )
ln(x + 1) = x − [(x2)/2] + [(x3)/3] − [(x4)/4] + ... = ∑n = 0 [(( − 1)nxn)/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.

### 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.

• Intro 0:00
• Taylor and Maclaurin Series 0:08
• Taylor Series
• Maclaurin Series
• Taylor Polynomial 1:20
• Lecture Example 1 2:35
• Lecture Example 2 6:51
• Lecture Example 3 11:38
• Lecture Example 4 17:29

### Transcription: Taylor Series and Maclaurin Series

Here we were asked to find the example, to find the McLauren series for ex.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 fn(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

fn(x), well ex, that is very easy, that is ex all the way down.0038

So, fn(0), that is just plugging 0 into ex, 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 fn(a)/n!/x-an.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! × x2 × 1/3! x3, and so on.0099

So, we get the McLauren series for ex is the sum from 0 to infinity of just xn/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 ex, is just xx/n!.0164

OK, for this example we are asked to find the McLauren series for f(x) = (ex)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) = ex. We are going to write down n fn(x). Oops, f(x) = (ex)2.0012

So f0(x) is just (ex)2.0031

First derivative is just (ex)2 × 2x.0036

The second derivative is 2x × derivative of (ex)2, so that is 2x × 2x × (ex)2.0041

Plus (ex)2 × derivative of 2x that is 2 (ex)2.0053

That can simplify down to 4x2 (ex)2,0062

I guess 4x2 + 2, we could combine those terms (ex)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 ex.0114

ex was the sum from 0 to infinity of xn/n!.0121

A much easier way than writing down all of these derivatives is to substitute in x2 for x.0136

So (ex)2 is the sum form n=0 to infinity of x2n/n!.0145

You could write that as the sum from n=0 to infinity of x2n/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 +x2/1.0176

n=2 gives you x4/2!.0182

n=3 gives you x6/3!, and so on.0187

Notice that that is just what you would have gotten if you had looked at the series for ex.0193

1+x+x2/2!+x3/3!.0199

If you had taken that series and just changed x into x2, 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 (ex)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 ex, and when you have to find (ex)2 you just do a quick substitution, and we get our series for (ex)20271

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 nth derivative there.0023

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

We will call that cn, that is the nth coefficient of the Taylor Series.0047

Then the x-an 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 tk(x), and what that means is, essentially that you get this thing that only runs up to xk 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 xk 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 nth derivative of f at a.0163

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

fN(0)/n! × normally x-an, but since a=0 it is just xn.0174

I want to figure out what these coefficients are, the cn's.0184

I will make a little chart here.0188

n and the nth derivative of x, then I will plug in the nth derivative of 0.0191

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

That is the same as fn(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 × x0 + 0 × x1.0302

So I will leave that out, - 1/2, 1/2! × x2, + 0 x3, + 1/4! × x4.0307

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 x5 - 1/6! × x6,0333

0 x7 + 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 x2n/2n!.0358

I am going to write -1n 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 - x2/2! + x4/4! - x6/6! 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 x2/2! is just x. Sorry 2x/2!0488

+ now the derivative of x4 + 4x3/4!.0503

- derivative of x6 is - x5/6! 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, 4x3/4!.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 x5, 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 x5/5! 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

x7/7!, 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 x2n+1/(2n+1)!.0601

And, -1n 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 ones0645

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 t4(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

n fn(x), the nth derivative, fn(pi/3).0718

Then the full coefficient cn, which is fn(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 nth derivative x, of a, you plug in x=a,0742

Divided by n! multiplied by the power part x-an.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 cn 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-a0 which is just 0.0924

n=1 says we take the 1/2 × x-a is pi/3, so pi/31.0931

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/33 + sqrt(3)/48 × x-pi/34.0967

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-an,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 nth derivative of f at x over n! × x-an.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-x2/2+x4/4! 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 ex 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-x2/2 + x4/24. 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 + 0x2 + 0x4, 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 - x2/2 + x4/24,1214

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

So that is + x2/2, then we multiply across there, x2/2.1228

x2/2 × -x2/2 is -x4/4 + x6/48.1237

We subtract again and the x2/2's cancel.1258

x4/4, I can write that as 6/24.1263

So, we have 6/24 - 1/24, that gives us 5/24 × x4 - x6/481269

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 x4.1292

sec(x) = 1+x2/2 + 5x4/24.1305

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