Finding function from power series by integrating | Series | AP Calculus BC | Khan Academy

We know that for (x) in the open interval from (-\frac{1}{2}) to (\frac{1}{2}), that (-\frac{2}{1-2x}) is equal to this series, and I say using this fact, find the function that corresponds to the following series. And like always, pause this video and see if you can work through it.

All right, so the first thing you might say is, well, how do we know that this expression is equal to this series? And you might recognize the series as a geometric series where the first term is (-2), and then the common ratio to get each successive term— we're multiplying, we're multiplying by (2x). We're multiplying by (2x). And so for a geometric series like this, the sum is going to be the first term, (-2), over (1) minus the common ratio, which is exactly what we have over there. So that's why we know that.

And this right over here gives us our radius of convergence—the (x) values for which this thing will actually converge. But now that we feel good about this first statement, let's try to answer their actual question. So we want to find the function that corresponds to the following series.

So my instinct is to say, well, how does this series relate to the series they gave us? Let's see. This first term is (-2), this first term is (-2x), this is (-4x), this is (-2x^2). So the thing that might jump out at you is that this second series they gave us is the anti-derivative of this first one. Or we could say this first series is the derivative of the second one. What's the derivative of (-2x) with respect to (x)? Well, it's (-2). What's the derivative of (-2x^2) with respect to (x)? Well, it's (-4x), and so on and so forth.

And so if we were to call this— let me call this thing right over here— actually let me just say that this is equal to (g(x)). Well, the way I can take this right-hand side and get (g(x)) is by taking the anti-derivative, taking the indefinite integral.

And so what we can do: we can take the indefinite integral of both sides (dx). So (dx) on the right-hand side, I'm going to get (g(x)), and on the left-hand side, I am going to get— actually let me write it this way— so on the left-hand side, I'll just rewrite it. I'm going to get the indefinite integral of— I'll write it as (-2dx) over (1 - 2x)— that's just another way of writing what I have on the left-hand side here. This is equal to (g(x)), right? If I take the antiderivative of this, or an antiderivative of this is (g(x)), where the constant would have been zero.

So I'll say is equal to (g(x)). And so the key here is, well, what's the indefinite integral of this stuff? And you might immediately recognize, well, this— what I have here in the bottom, I have its derivative here on the top. If I consider this to be (u), if I say (u = 1 - 2x), this is just (u) substitution. Then (du) is equal to the derivative of this with respect to (x), which is (-2dx), and so I have the (du) right over here.

So let me rewrite all of this. I can rewrite this as the integral of (du) over (u) is equal to (g(x)). And this we can rewrite as the natural logarithm of the absolute value of (u), plus (c), is equal to (g(x)). And then we can undo our (u) substitution. And so for (u), I'll undo— I will substitute back the (1 - 2x), so I can write the natural logarithm of the absolute value of (1 - 2x) is equal to (g + c).

I want to forget that (+ c) is equal to (g(x)). And so the next thing we want to do is, well, what is our (c)? And the easiest way to figure out (c) is let's substitute (0) onto both for (x). And so let's think about this a little bit. So if I put a (0) here, if I put (x = 0)— actually, let me just write it again— so I'll have the natural logarithm of— if I say (x) is (0), (0), this is going to be the natural logarithm of the absolute value of (1), plus (c), is equal to (g(0)).

So what's (g(0))? (g(0)), every one of these terms is equal to (0). (g(0)) is equal to (0). So is equal to (0). Well, the natural logarithm of one is just (0), so (0 + c) is equal to (0). (c) is equal to (0).

So there you have it. I just took the antiderivative of both sides of this equation right over here. I figured out when I substituted (0) for (x), and okay, my constant here is going to be (0), and I get that this series, this (g(x)), is equal to the natural logarithm of the absolute value of (1 - 2x).

Now, if we keep this restriction that we're in this open interval, then this is always going to be positive, and we wouldn't have to write the absolute value. But this— we could be safe by writing the absolute value. But there you go, using the fact above, we found the function that corresponds to this following series.

And I guess you could say the trick of it was recognizing that the series up here is the derivative of the series down here. And so this is going to be the derivative of the following series. And so we just took the anti-derivative of both sides or the indefinite integral of both sides.