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

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

Integrate. [SOUND] We've already seen how we can differentiate term by term. Well can we integrate term by term? Well yes, and here's a theorem. Suppose I got some function f, and f of x is this power series, the sum n goes from zero to infinity of a sub n x to the n. And big R is the radius of convergence of this power series.

Well then, I can integrate this function term by term, meaning that the integral of f of x dx, x goes from zero to some parameter t, is the sum n goes from zero to infinity of this, a sub n times t to the n plus one over n plus one. And this is true for any value of t between minus r and r. If you want to be a little more specific, it turns out the radius of convergence of this series is exactly the same is exactly the same as the radius of convergence of this series.

Well, and if you're wondering where this term comes from, right, that is exactly what you would get just by integrating the term a sub n x to the n, right? The integral of a sub n x to the n dx, x goes from zero to t, well that just is a sub n times t to the n plus one over n plus one. Armed with this result, we can do some pretty great stuff. For example, we know that the sum n goes from zero to infinity of x to the n, well, that's one over one minus x, as long as the absolute value of x is less than one. I mean, in other words, right, in this particular instance, the radius of convergence, big R, is one. Well, let's integrate that. So, the integral, from x equals zero to t of one over one minus x d x. Yeah, what is that interval? Well, I can evaluate that by making u substitution, so let's set u equal to one minus x, and in that case, d u is just minus d x. I don't see a minus dx there, but I can artificially create a minus dx with some carefully placed minus signs there. Okay, now I've got a negative dx there. That looks great. So, this is negative the integral, x goes from zero to t of, I've just got a du over one minus x, but that's u. So I want to know, how do I anti-differentiate 1 over u. Well this is negative log the absolute value of U. Then evaluate it at x equals t and x equals zero. Now take the difference, so that's negative log of well what's U, absolute value, one minus x, and we are evaluating this at x equals zero and at t. So this is negative log of the absolute value of one minus t. And then it would be subtracting this with x equals zero, but I'm subtracting anegative, so I'm going to add the log of the absolute value of one minus zero. Well, what's log of one? Log of one is just zero, so I don't need to include this term. So all told, if I integrate one over one minus x from zreo to t, what I'm getting is negative the natural log, the absolute value of one minus t. Now we can integrate the power series term by term. Alright, so we just showed that negative log the absolute value of one minus t is the integral x goes from zero to t of one over one minus x d x. Now I'm going to replace this one over one minus x with a power series, so this is equal to the integral, x goes from zero to t instead of one over one minus x, the sum n goes from zero to infinity of x to the n, because one over one minus x is equal to this. Well, as long as the absolute value of x is less than 1 d x. Okay. Now I'm integrating a power series, and by the theorem, I can do that term by term. So this is the sum, n goes from zero to infinity of the integral of x going from zero to t, of x to the n d x. So now I have to be able to do this integration problem. So this the sum, n goes from zero to infinity of, what's this? Well, I'm integrating x to the n d x from x equals zero to t. So that's t to the n plus one over n plus one. So there it is, right? I found a series for this function, for negative the natural log of the absoulte value of one minus t. Here's a power series for that function, and it's valid for which values of t? Well as long as t is less than 1 in absolute value. Before we get too excited, it's worth mentioning a fine point here. Well one case when you'd like to use this formula is when t equals negative one. Because in that case, what do you get? Well, then you get negative the natural log of the absolute value of 1 minus negative 1, is equal to what? Well, it's equal to this series, the sum n goes from zero to infinity of, now, negative one to the n plus one over n plus one. I can simplify this a bit. Right, what is this? Negative the natural log of one minus negative one, well this is just two. So this is negative the natural log of 2 is given by this series. Or is it? Alright the issue here is that I'm going to plug in t equals negative one. But strictly speaking, that's not one of the values of t that I'm allowed to plug in. If you want to dig deeper, there are some other theorems that you might want to look into. For example, it turns out that this is true. Alright, it turns out that negative the natural log of 2 is given by this alternating series. But to justify that, you do a little bit more. One way to do it is by invoking Abel's Theorem. [SOUND]

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