Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

898 ratings

The Ohio State University

898 ratings

Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Power Series

In this fifth module, we study power series. Up until now, we had been considering series one at a time; with power series, we are considering a whole family of series which depend on a parameter x. They are like polynomials, so they are easy to work with. And yet, lots of functions we care about, like e^x, can be represented as power series, so power series bring the relaxed atmosphere of polynomials to the trickier realm of functions like e^x.

- Jim Fowler, PhDProfessor

Mathematics

Transform! [MUSIC] We have some series that we already know about. Well, for example I know that the sum n goes from 0 to infinity of x to the n. That's a power series for a function we already know. Its just the power series for one over one minus x. And we've got some operations that we can apply to power series. Well, we can differentiate a power series, we can integrate a power series term by term, I can multiply together a power series and I can divide power series. Now we can put together the series that we know and the operations that we can do in order to build new power series. Well, here's our goal.

I'd like to find a power series for the function 1/(1-x)^2. We can do this in two different ways. Let's notice something about this. This e is the derivative of 1/(1-x). Why is that? Well, let's just work it out right, this is the derivative I'm claiming of 1 minus x to the minus 1st power, well differentiate this using the power rule that's negative 1 times 1 minus x to the negative 2nd power times the derivative of the inside by the chain rule which is just multiplying this by another copy of minus 1. So this is minus 1, times 1 minus x to the negative second power, times minus 1. Well, what happens, right, these minus 1s cancel. And what I'm left with, is just 1 minus x to the minus second power. Well, I could write that as 1 over 1 minus x squared. That's exactly what I've got here. So the derivative of 1 over 1 minus x is this function that I'm interested in? So it's differentiate the power series term by term. So this is the derivative of the sum n goes from 0 to infinity of x to the n. And, like I said, differentiating term by term. This is the sum n goes from 1 to infinity. Of the derivative of nth term the derivative of x to the n, note that I changed from n equals 0, n equals 1 because when derivative the n equals 0 terms that's the derivative of a constant which is just 0. So when I start here with n equals 1, alright so this is the sum and it goes from 1 to infinity what's the derivative of x to the n by the power rule that's n times x to the n minus 1. And now if I don't like having the n minus 1 there I could re-index this series, right. What happens when I plug in n equals 1? Well, that's 1 times x to the 0. Right. And then what happens when I plug in n equals 2? Well, that's 2 times x to the 1st. What happens when I plug in n equals 3? That's 3 times x squared, and so on. So I could rewrite this series as just the sum n goes from 0 to infinity of (n+1)*x^n. And there we go, I've written down a power series for the function 1/(1-x)^2. Of course, if I don't like thinking like this, if I don't like differentiating, I can also approach this problem by just trying to multiply together two copies of the original power series. Well, what I mean, is that the sum of x^n n goes from zero to infinity is 1/(1-x).

That means that if I square this side and square this side I should get a formula then, a power series for 1 over 1 minus x quantity squared. But how do I write that as a power series? Well, lets at least get started, lets at least figure out what the first few terms are. So I'm just going to write down the sum of x to the n or you know the first few terms of it. So that's 1 plus x plus x squared plus x cubed plus... and I'm squaring that so multiplying it by itself so I'll write down the same thing again, one plus x plus x squared, plus x cubed plus... Now I want to figure out what happens when I multiply these things together. So what do I get? Well, 1 times 1, thats 1.

How many x terms do I get here? Well, I've got a 1 times and x, or I've got an x times a 1. Now there's no other way to multiply these things and get and x. So theres plus 2 x.

How many x squareds do I get out? Well, I can multiply this 1 times this x squared, multiply this x times this x or I can multiply this x squared times this 1 but I got 3x squared and there's no other way to get an x squared out. how can I get an x cubed? Well, I can multiply 1 times x cubed. Or x times x squared or this x squared times this x or this x cubed times 1, so that gives me 4 ways of getting an x cubed term and alright plus... Now, maybe you believe that pattern continues. Well, it certainly looks like this is giving me the sum n goes from 0 to infinity of n plus 1 times x to the nth power. I mean, n equals 0 is 1 times x to the 0, n equals 1 is 2 times x to the first, n equals 2 is 3 times x squared, n equals 3 is 4 times x cubed and so on. To make that more rigorous, we probably have to talk about induction. Or we could just bring up a theorem on multiplying power series. Well, here's out theorem for multiplying power series. The product of this power series and this power series is given by this power series. And it's a little bit complicated to see how the coefficient's affected, right? Here the coefficient for a suborn, here the coefficients for b suborn and when I multiple these together I get this concolved series the sum i goes from 0 to n of a suborn times b suborn n minus i. Now we can apply that theorem, well in this case our formula for 1 over 1 minus x is just for all the coefficients are 1. So if I multiply 1 over 1 minus x by 1 over 1 minus x, then I get 1 over 1 minus x squared. So that means all the a sub i's and b sub n minus i's are all just 1, so here is a power series for 1 over 1 minus x squared. Lets simply, well the sum i goes from 0 to n of just 1 or that's 1 plus 1 plus 1 but its n plus 1 once. So I can really write this power series which is the sum n goes to 0 from infinity of n plus 1 times x to the n and that's exactly what we got before by using derivatives. [SOUND]

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