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Lecture Comments (14)

1 answer

Last reply by: Dr. William Murray
Thu Aug 4, 2016 6:03 PM

Post by Peter Ke on July 31 at 02:05:04 PM

For example 3, how did you know that the interval of convergence is from -1 to 1?

1 answer

Last reply by: Dr. William Murray
Thu Jun 9, 2016 1:23 PM

Post by Silvia Gonzalez on June 9 at 09:20:29 AM

Thank you again for the class. I have a question about Additional Example 4. When you test for convergence at x=8 (upper limit of the interval of convergence) you use the Limit Comparison Test comparing with 1/n^2. Could I have used the Comparison test (which is easier) because 1/n^2 is always bigger than the original series and so they would both converge, or is there a reason why it can not be used in this case? Thank you.

3 answers

Last reply by: Dr. William Murray
Sat Apr 25, 2015 7:32 PM

Post by Luvivia Chang on April 18, 2015

Hello Dr William Murray
I love your lectures very much. And you make them really fun to watch. I have a question here. Since we have learned a few ways of tests to determine divergent or convergent, what is the purpose of doing so? What is the significance of determining this or any application?
Thank you.

1 answer

Last reply by: Dr. William Murray
Thu Apr 10, 2014 7:50 PM

Post by Brandyn Albrecht on April 7, 2014

In example 4, you said that (-6)^n/6^n= (-1)^n, shouldn't it just be -1?

3 answers

Last reply by: Dr. William Murray
Mon Nov 12, 2012 12:44 PM

Post by Paul Carrera on April 21, 2011

I love the series lectures. Thank you for making them easy to understand. Do you have any lecture videos on Representation of Functions as a Power Series? I don't see these videos in this syllabusss, and I need help with them. Thanks

Power Series

Main definitions and pattern:

Definitions: A power series is a series of the form . (The cn ’s are the coefficients, expressions that might involve n, but won’t involve x.)

Pattern: The power series always converges for values of x within some radius R around the center a. For a − R < x < a + R (i.e. |x − a| < R), it is Absolutely Convergent . For x < a − R or x > a + R (i.e. |x − a| > R), it Diverges . At the endpoints x = a − R and x = a + R, it might be conditionally convergent, absolutely convergent, or divergent.

R is called the radius of convergence. We can have R = 0 or R = ∞. The interval a − R < x < a + R (or a − Rx < a + R, or a − R < xa + R, or a − Rxa + R) is called the interval of convergence.

Hints and tips:

  • For most power series, you can use the Ratio Test to find the radus of convergence.

  • On a few examples, you should use the Root Test. These examples usually have the form (a function of n)n .

  • However, you can never use the Ratio or Root Test to check the endpoints, since they will give you L = 1, which is inconclusive.

  • You must always check each endpoint individually, using some test other than Ratio or Root. Common favorites are the Limit Comparison Test with a p-series, Alternating Series Test, and Test For Divergence.

  • A factorial in the denominator often leads to R = ∞.

  • Memorize the geometric series expansion for -1 < x < 1.

  • You can often derive other power series from the geometric series by the following methods.

    • Algebraic manipulations, e.g. multiplying by x. These won’t change the radius of convergence or whether the series converges at the endpoints.

    • Substitutions, e.g. replacing x by 2x or x² . This will change the radius of convergence.

    • Derivatives and integrals. These won’t change the radius of convergence, but they might change whether the series converges at the endpoints.

  • Other common series that are worth memorizing (although they can be derived from the geometric series) are for −1 ≤ x < 1 and arctan for −1 ≤ x ≤ 1.

Power 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
  • Main Definitions and Pattern 0:07
    • What Is The Point
    • Radius of Convergence Pattern
    • Interval of Convergence
  • Lecture Example 1 3:24
  • Lecture Example 2 10:55
  • Lecture Example 3 14:44
  • Additional Example 4
  • Additional Example 5

Transcription: Power Series

OK, here we are working with some more examples of power series.0000

In particular, finding the intervals of convergence.0004

Here I have one that looks very complicated.0008

x-2n/6n × a big polynomial.0012

So again, we will start out with a ratio test.0015

The ratio test says that you look at the n+1 term divided by the n term.0019

So I get x-2n+1/6n+1, and then this big polynomial n+12 + 3 × n+1 - 7.0027

All of that divided by the an term, so x-2n.0048

Divided by 6n, and n2 + 3n - 7.0054

That is large and messy, but what I am going to do is I am going to flip the denominator,0062

And pair off the factors in the denominator with their similar factors in the numerator.0070

I will get x-2n+1/x-2n.0078

6n in the numerator, 6n+1 in the denominator.0089

In the numerator I get n2 + smaller stuff.0102

In the denominator I get n+12 + smaller stuff.0109

I still need absolute values at least on the x part.0116

In turn, that simplifies down the x-2, everything cancels there but x-2, in absolute value.0123

6n/6n+1 cancels down to 6 in the denominator.0134

n2/n+12, that is a big polynomial, but both of the big polynomials,0142

Their leading term is n2, so if you divide both terms by n2,0150

What happens there is that the n2 terms will go to 1 and everything else will have an n in the denominator,0159

The whole thing will go to 1. Just 1/1.0166

Remember the whole point of doing the ratio test is that it responds very well to the geometric terms.0171

That is x-2n/6n.0178

It gives you a really nice answer with those terms.0181

When you do the ratio test on a polynomial, it just gives you 1 when you take the limit0184

This polynomial, this big polynomial, the ratio test kind of sweeps it away and kind of makes it into a 1.0190

Our answer, when we take the ratio test is just the absolute value of x-2/6.0201

The ratio test says that whatever you get for your l,0207

That should be less than 1 for the thing to converge.0213

So now I just have a little bit of algebra to figure out which values of x make that work.0216

I can solve that into x-2 < 6, so x-2 < 6 and less than -6.0220

So, x is between 8 and -4 there.0238

That is the region within which the series is absolutely convergent. Outside that region it is divergent.0247

And remember, the endpoints we never know, we always have to check those endpoints individually.0256

Let us check those endpoints.0263

We will check those endpoints individually.0276

The first is x=-4, so if you plug in -4, you get -4 - 2n.0278

So that is -6n/6n.0287

Then, the n2 term.0293

The -6n and 6n gives you -1n/n2 term.0300

That is a series that we know converges because it is an alternating series and it converges by the alternating series test.0311

That was checking the left endpoint, x=-4.0327

x=8, that gives us the series, ok 8-2n is 6n/6n × the n2 term.0332

That simplifies down to 1/the n2 term.0351

That is a series that is very similar to 1/n2.0362

We can use the limit comparison test with the sum of 1/n2.0369

I will not work out the details of that.0378

You divide the two series together, you calculate the limit, the limit is going to turn out to be 1.0382

The important thing here is that 1/n2 is a p series.0388

With p=2. That is convergent.0395

The limit comparison test says whatever the new series does, the old series must have done too.0405

That one converges as well.0415

Both endpoints in this case turned out to give us convergent series.0420

The interval, our final answer here, is x goes between -4 and 8,0426

But since both endpoints make it converge, I am going to put equal signs on this.0436

We include both endpoints there.0441

If you write that in interval notation, it is -4 to 8, with both endpoints included there.0443

Let us recap there.0455

Again, power series we worked with the ratio test, so we worked out the big ratio.0458

It looks really nasty but remember the point of the ratio test was it kind of reduces you down to geometric terms.0463

The polynomial terms all get swept away to 1.0472

So, the polynomial terms get swept away, we are reduced down to x-2/6 < 1.0476

Do a little algebra to simplify that.0483

Then we just have to check whether these two endpoints are included or not.0487

The endpoints, remember, you cannot use the ratio test.0491

Because the ratio test, if you did use it, it would always give you 1.0495

Because those are exactly the cutoff points where the ratio test gives you 1.0499

You cannot use the ratio test, so for each endpoint we plug it in separately and use some other test,0505

It turns out that both of those do converge.0512

So, when giving the answers, we include both endpoints there.0516

One final example here is we want to find the function arctan(x) as a power series0000

Centered at 0, and find the interval of convergence of that.0007

Again, the trick here is to think about derivatives and integrals and remember that the derivative of arctan(x) is 1/1+x2.0012

Then, that is something that we can write as 1/1-(-x2).0028

The point of that is to make it look like the sum of a geometric series.0038

That is equal to the sum from n=0 to infinity of -x(2)n0042

The easiest way to think about that is probably to work backwards.0054

Remember that its sum will be exactly given by that formula.0058

That is for -x2 in absolute value less than 1.0063

Motivated by that, we will write 1/1+x2.0071

I am going to expand out this series in two individual terms.0079

That n=0 gives you 1, - x2 + (x2)2, so that is x4 - x6, and so on.0085

This is for x2 in absolute value less than 1.0100

That is x is between -1 and 1 there.0108

Arctan(x) is the integral of 1/1+x2.0118

I am going to integrate both sides here and I get, the integral of this, taking the integral of the right hand side, is x - -x3/3 - x5/5 - x7/7,0128

We still have the problem of adding on that constant, so I will add it on at the beginning this time.0155

C + x + x3, and so on.0162

To find the value of C I am going to plug in x=0,0166

So I get arctan(0) = C + 0 + a bunch of zeroes.0172

arctan(0) = 0, so I get C = 0, so again I do not need to worry about my constant here.0182

I get arctan(x) = x - x3/3 + x5/5 - x7/7 and so on.0190

If you want to write that in a nice concise form, you can write that as the sum from n=0 to infinity,0208

Now I just want to catch the odd powers here,0216

So, a way to catch the odd powers is to write x2n+1/2n+1.0220

That will just give you the odd values.0235

To make it alternate in sign, I am going to put a negative 1 to the end here.0238

So, there is a power series for arctan(x).0247

Now, we still have to find the interval convergence.0252

What we know is that the original series converged for x between -1 and 1.0257

That means the new series, after we take its integral, is also going to converge for x between -1 and 1.0264

But we still have to check the endpoints,0271

Because when you take the derivative of the integral, that can change the convergence at the endpoints.0272

So we check the endpoints, always have to check them separately.0279

So x=-1, gives us, if we plug that into this series, -1.0289

Now, -x3, that is - -13/3 + -15/3 - -17/3,0301

So that is -13 is -1, and then there is one more negative, so this is plus 1/3,0322

Oops, I should not have put 3's in all the denominators, those should be 5 and 7.0336

So, minus 1/5.0344

So 7 -1's and one more -1, gives you + 1/7 and so on.0345

This is an alternating series.0354

The terms go to 0 and they get steadily smaller.0357

They are decreasing so this converges by the alternating series test.0361

If we check x=1, the other endpoint, plug that in and we get,0370

1 - 1/3, plugging that into the series, + 1/5, - 1/7, and so on.0375

That is also an alternating series and it also satisfies the condition of the alternating series test.0385

So this converges by the alternating series test.0392

So this time both endpoints converge so our interval is x between -1 and 1.0398

Both endpoints are included, or if you write it in interval notation, you use square brackets on both of those, form -1 to 1.0410

Again, the trick there for finding a power series for arctan(x),0420

Was to think about the derivative, 1/1+x2,0424

And that can be written in a form that lets you convert it into a geometric series.0429

Once it is a geometric series, we can undo the derivative by taking its integral and get back to a series for arctan(x).0435

To find the integral of convergence, well we just use the old interval of convergence.0445

Then, because we took the integral, we have to recheck the endpoints of the series,0451

So we check each one of those endpoints separately, they each converge,0455

And so we throw each of those endpoints into the intervals. 0460

Thanks for watching, this is educator.com0464

Hi, this is Will Murray for and we are here to talk about power series.0000

A power series is a series of the form, it has a coefficient cn,0007

Then it has a power of either xn or x-an.0014

The game with power series is you are trying to plug in different values of x and see what values of x.0020

When you plug in a value of x, you get a series just of constants.0030

Then the question is which values of x make that thing converge.0035

If you try out some examples, we will work on some examples soon.0040

You start to notice a pattern which is that you always have a center value at a.0045

Now you notice that if you plug in x=a here, then you get a-an, so you just get a series of 0's, that always converges.0055

It always converges at x=a.0064

It turns out that it always converges in some radius around that center.0067

So, there is always some radius r around that center.0072

And the series always converges absolutely between those values.0080

That is a+r, and this is a-r.0085

It always converges absolutely in that region.0091

That region is absolute convergence.0098

It is always divergent outside that region.0100

Then at the two end points, a-r and a+r, for those values of x, it is really unpredictable what the series does.0105

Sometimes it will diverge at both of them, it could converge at both of them.0114

It could converge at one and diverge at the other one.0120

There is no way to predict it ahead of time, just every particular example, you have to check each one of those endpoints separately.0125

To recap here, it always converges between a-r and a+r.0132

Another way of saying that is that x-a, in absolute value, is less than r.0138

Because it is saying x is within r units of a.0146

r is called the radius of convergence, it can happen that the radius is 0 or infinite.0152

We will see some examples of that.0158

Then the interval, what you can do is you can look at the interval from a-r to a+r.0162

That interval is called the interval of convergence.0172

That interval might include both endpoints or not include both endpoints, or it might include one of them and exclude the other one.0178

There are all of these different possibilities about whether the endpoints are included.0188

That interval, including whichever endpoint makes it converge is called the interval of convergence.0194

That is a lot of definitions to start with.0203

It will be easier once we look at some examples.0206

Here is an example, we want to look at the interval of convergence for the power series xn/2n × n.0208

The trick with power series is you almost always want to use the ratio test.0216

You almost always want to start out using the ratio test.0223

We will start out using the ratio test.0229

Remember the ratio test says you look at an+1/an in absolute value.0231

That in this case is xn+1/2n+1 × n+1/xn/2n × n.0238

All in absolute value.0254

I will flip the denominator up and multiply it by the numerator.0255

I am going to organize the terms with the terms that look like them,0259

We get xn+1/xn.0264

2n/2n+1, and n/n+1.0269

I cannot get rid of the absolute values completely because the x could be positive or negative.0280

This simplifies down into the absolute value of x.0287

n/n+1, when you take the limit as n goes to infinity, that just goes to 1.0291

Then 2n/2n+1 just gives you a 2.0298

Remember, the ratio test says it converges absolutely whenever that limit is less than 1.0304

We look at abs(x)/2 < 1, and that is the same as saying the abs(x) < 2.0313

So, x is between -2 and 2.0326

Those are the values of l that give you a value less than 1.0333

Those values of x make this series absolutely convergent.0340

If x < -2, or x > 2, we know that the ratio will be greater than 1.0344

The series will be divergent.0355

Let me fill in what we have learned so far.0358

From -2 to 2, we know that it is absolutely convergent.0361

If it is less than -2, we know it is divergent,0372

Or if x is bigger than 2, we know it is divergent.0381

The question we have not answered yet is what happens at the endpoints.0383

A big rule here is you cannot use the ratio test at the endpoints.0387

The reason is the endpoints are exactly where the ratio test gives you 1.0393

The ratio test, when it gives you 1, tells you nothing.0398

You have got to use some other test at the endpoints.0403

Let us check the endpoints.0408

Without using the ratio test because I know if I use the ratio test, I am just going to get 1.0413

Let us check these separately.0423

If x=-2, then our series turns into -2n/2n × n, which simplifies down into -1n/n.0425

That is a series that we have investigated before.0442

This converges by the alternating series test.0448

That is the alternating harmonic series, and the alternating series test applies to it.0454

x=2, gives us the series of 2n/2n × n, which gives us the series the sum of 1/n.0460

That is the harmonic series, or if you like, that is a p series with p=1 and we know that diverges.0477

We have seen that one before.0481

That is the harmonic series.0486

You can also think of it as a p series with p=1.0491

The important thing there is that 1 is less than or equal to 1.0498

Anything < or = 1 makes it diverge.0501

So x=-2 makes it converge, x=2 makes it diverge.0508

So, the interval of convergence is everything between -2 and 2.0512

And we include -2 because it makes it converge.0533

We exclude 2 because it makes it diverge.0536

Another way of writing that in interval notation is to say the interval from -2 to 2,0540

You put square brackets on the -2, that means it is included,0546

Round brackets on the 2, that means it is excluded.0551

That is your final answer there.0558

Let me recap here.0560

The important way to approach power series, most of the time you want to start out with the ratio test.0563

At least 95 times out of 100, you want to start out with the ratio test.0571

You work through the ratio test and you get your limit.0577

You set that less than 1, because remember less than 1 is what the ratio test looks for to tell you the series converges.0581

That tells you the basic interval except it does not tell you the endpoints.0592

It tells you that it is absolutely convergent between those points, divergent outside those points.0597

Then you check those two endpoints separately.0605

At first you do not know what the endpoints do, and remember, you cannot use the ratio test for the endpoints, because the ratio test will give you 1.0610

When the ratio test gives you 1, you get no information and you have to try something else -- you have to plug them in individually.0619

x=-2, it turns out that that one works,0627

x=2, it turns out that it makes it diverge.0631

When we are writing the interval we include x=-2, and exclude x=2, and if we write that in interval notation, we put square brackets on -2, and round brackets on 2.0637

Let us try another one.0654

We want to find the interval of convergence for the power series of -1n × xn/n!.0656

Again, we will start with the ratio test.0663

We will look at an+1/an.0667

I know that these -1's will get swept away by the absolute values anyway, so I am not even going to write the -1 terms.0672

I am just going to write xn+1/n+1! divided by the an term, xn/n!.0680

Still need my absolute values here.0694

If I flip my denominator and pull it up, we get xn+1/xn × n!/n+1!.0697

That simplifies down to x abs(x).0711

Now n+1!, remember you can write that as n! × n+1.0718

So, the n!'s cancel and we are just left with x/n+1.0725

Remember, we take the limit of this as n goes to infinity.0733

We are plugging in different values of x here, but whatever value of x we plug in,0738

It is a constant. n is the one going to infinity.0747

When n goes to infinity, no matter what value of x you start with, this thing goes to 0.0752

For all values of x, no matter what value of x you start with.0764

This thing goes to 0 because n is the one going to infinity.0770

So, l is 0 no matter what you start with.0776

The key point here is l < 1, so the ratio test says that no matter what value of x you put in there, it is absolutely convergent.0778

For all possible values of x, for all real numbers of x.0810

If you want to write that in interval notation, the answer would be (-infinity, infinity).0816

I put round brackets there because there is no question of endpoints here because infinity and -infinity are not actual numbers we can plug in there.0824

There is no question of plugging in endpoints or considering endpoints to be included or excluded.0833

There are no endpoints.0840

We say that the interval runs over all real numbers from -infinity to infinity.0845

Another way of saying that would be that x going from -infinity to infinity.0853

We started here by applying the ratio test.0861

We wrote down the ratio, we took the limit as n goes to infinity, because that 0, which < 1, no matter what value of x it is, we get that it converges for all possible values of x.0866

Our third example here, we are not given a power series, we have to write a power series ourselves.0882

For ln(1-x).0891

Then we have to find the interval convergence.0892

This one is not so obvious how to start unless you have seen something like this before.0895

The key point is to remember that the derivative of ln(1-x),0901

Is 1/1-x × derivative of 1/x, so that is -1/1-x.0914

1/1-x is exactly the sum of a geometric series.0926

That is the sum from n=0 to infinity of xn, put a negative there.0932

That is a geometric series and that is true for abs(x) < 1.0943

That is probably easier to think about if you work backwards.0952

If you look at this geometric series, that sums up to the first term/1 - the common ratio,0955

So that is equal to 1/1-x.0964

Armed with that intuition, we can write ln(1-x) as the integral of 1/1-x dx.0970

Except there was that negative sign, so I will put that on the outside.0981

That is the integral, now 1/1-x we said was this geometric series.0988

That is 1+x+x2, and so on, dx.0995

If you integrate that, you get negative, integral of 1 is x, integral of x is x2/2,1009

Integral of x2, is x3/3, and so on.1020

But this is not an indefinite integral, so we always have to include a constant.1024

How do we figure out what the constant should be.1032

To find the constant, we are going to plug in x=0 to both sides.1035

We get ln(1-0) = -, well if we plug in 0 to a bunch of x terms, we get a bunch of 0's + C.1043

So, the constant is ln(1), which is 0.1060

So that is nice, the constant disappears.1064

We get ln(1-x) is equal to, I will distribute the - sign, -x - x2/2 - x3/3, and so on.1067

We can write that as the sum from n=1 to infinity, of -xn/n.1087

Notice that that is what we would have gotten if we had integrated this geometric series directly.1105

If we took the integral of xn, we would have gotten,1111

Well, xn+1/n+1, then shifting the indices from n=0 to n=1,1114

Would have converted that into xn/n.1123

That is the power series.1129

What we now have to do is find the integral of convergence.1132

What we knew before, is that x was between 1 and -1.1137

When you take the derivative of integral of a power series, it does not change its radius of convergence.1144

We know it still goes from -1 to 1.1152

However, it might change what happens at the endpoints.1156

So, we still have to check the endpoints of this series.1162

Sorry, check the endpoints of the integral, and see whether it converges or diverges at those two endpoints.1171

Wee have to check those separately.1179

Let us try x=-1 first.1181

that would give us the series of -1n/n,1183

That is the alternating harmonic series.1190

We know that converges by the alternating series test.1193

x=1 gives us the sum of -, well 1n/n,1202

That is the negative of the harmonic series,1218

And we have seen several times that that diverges, either by just saying that that is a harmonic series,1220

Or by saying it is a p series, with p=1,1231

Since 1 < or = to 1, that makes it diverge.1235

x=1 makes it diverge, x=-1 makes it converge.1240

So, the interval, you could write that as x between -1 and I am including -1 because it made it converge,1245

I am not including 1 because it made it diverge.1256

In interval notation that is [-1,1),1260

With a straight bracket on -1 because it made it converge and a round bracket on 1 because it made it diverge.1265

Our answers there, we found the series by taking the derivative of ln(1-x),1275

We get something we can convert into a geometric series,1283

Then to get back to ln(1-x), we took that geometric series and we integrated it back up to get ln(1-x).1291

Now we have a power series for ln(1-x).1300

To get the interval, we used the interval for the geometric series, but then because we are taking derivatives and integrating,1303

We know that the endpoints, whether or not it converges at the endpoints, that could possibly change.1312

We have to check those again, we check -1, it converges.1318

1 diverges, and so we reflect those in the answers for the interval.1322

We will try some more examples later on.1330