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

899 ratings

The Ohio State University

899 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

Taylor Series

In this last module, we introduce Taylor series. Instead of starting with a power series and finding a nice description of the function it represents, we will start with a function, and try to find a power series for it. There is no guarantee of success! But incredibly, many of our favorite functions will have power series representations. Sometimes dreams come true. Like many dreams, much will be left unsaid. I hope this brief introduction to Taylor series whets your appetite to learn more calculus.

- Jim Fowler, PhDProfessor

Mathematics

Centered around A. [SOUND] Well thus far, we've been talking about Maclaurin series or about Taylor series centered around zero. But, I can write down a power series for a function, not centered around zero, but centered around some other point A. Well what I mean is I could try to write the function f as a power series. N goes from zero to infinity, some coefficient C sub n, not just times x to the n, but times x minus a to the nth power. Well, just like before, let's assume that f has a power series representation, and then try to figure out what those coefficients must be. I want to be a little bit more precise, right. What I'm saying here is I'm going to assume this holds whenever x is, say, within big R of a.

And assuming this is true, I want to figure out what these coefficients have to be in terms of the function, f. So I'm just going to write down the first few terms, right. When I plug in n equals 0, I just get C sub 0 times 1, because it's something to the 0th power, plus when I plug in n equals 1, I get C sub 1 times x minus a to the 1st power. Plus, when I plug in n equals 2, it's C sub 2 times x minus a squared. And then it keeps on going. Now, what happens when I plug in a for x? So that's asking, what is f of a? Well in that case it's C sub zero plus what's this term? It's just zero plus what's this term? Well it's just zero plus, all the other terms are zero. So this is telling me that C sub zero is equal to f of a. Or another way to say that, is if I knew that this was true for values of x within big R of a, then I know what C sub zero has to be. C sub zero has to be the value of f at the point a. The next coefficient can be computed in terms of the derivative, but not the derivative at zero, but the derivative at the point a. Well let's differentiate this, see what we get. So I'm assuming that this is true whenever x is within big R of a. So the same will be true for the derivative, at least, so the derivative of f of x will be the sum, n goes not from 0 but from 1 to infinity of the derivative of this, which is C sub n times n times x minus a to the n minus 1, and this is true at least when x is within that same big R of a. Okay, let me write down the first few terms here. Let's plug in n equals 1. I get just C sub 1 times 1 times x minus a to the 0th power. So that's just C sub 1, plus when I plug in n equals 2. I get C sub 2 times 2 times x minus a to the 1st power.

When I plug in n equals 3, I get C sub 3 times 3 times x minus a squared, and then it keeps on going. And now note what happens when I plug in a for x. Alright? What's f prime of a? Well it's just like what happened before! I've got C sub 1, but then this C sub 2 times 2 times x minus a, that term vanishes. This term here also has an x minus a, so when x is equal to a, this term vanishes, all the later terms vanish. This means that I can recover the value of the C sub 1 coefficient, just by differentiating the function at the point a. And so on, right. Just like before, I can now write down a formula for the nth coefficient. So in general, what I'll find is that C sub n can be computed by taking the nth derivative of f at the point a and dividing that by n factorial. So let me summarize all of this. The Taylor series for function f centered at a, or sometimes you'll see just the Taylor series around a, for the function f, it's given by this. It's the sum n goes from zero to infinity, of the nth derivative of f at a, that's the big difference, divided by n factorial. Here's another difference. x minus a to the nth power. I mean it sort of makes sense. Right. The Taylor series around a, is a power series around a. And so it's got this x minus a to the nth term. But instead of calculating the derivative at zero we're calculating the derivative at the point a. [NOISE]

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