 Welcome back, everyone. In this video, we're going to introduce the notion of what's called the binomial series, which is a very important Maclaurin series that represents a large family of functions. And so let's consider it here. We want to find a Taylor series centered at zero, aka a Maclaurin series, and the function and consideration is f of x equals 1 plus x to the k, where k is some, is any real number, but it's a fixed number, won't change throughout this process. Now, the reason we're going to focus on this function right here is that if we're going to have any chance of describing a Maclaurin series, the derivatives have to be somewhat predictable. And the nice thing about 1 plus x to the k is by the power rule, we can predict what those derivatives are going to be. So if you take the 0th derivative, that's just the function itself, so you get a 1 plus x to the k. If you take the derivative by the power rule, you'll get k times 1 plus x to the k minus 1. It goes down. If you take the second derivative, you're going to get k times k minus 1 times x plus, or 1 plus x to the k minus 2, the power goes down. Then if you do the third derivative, you get a k from before, k minus 1 from before. Now you get a k plus, or k minus 2 from the previous power, and then you also times that by x, 1 plus x to the k minus 3. Like so. And if you do this over and over and over again by mathematical induction, we can see that the nth derivative of f will look like k from the first derivative, k minus 1 from the second derivative, k minus 2 from the third derivative, all the way up to k minus n plus 1 from the n minus first derivative, and then you're going to have a 1 plus x to the k minus n power. So we started off with k, and now we've subtracted n from it because we've done n derivatives. We see this pattern that follows by induction. Now for Taylor's equation, we see that the coefficient sequence inside of a Taylor series, it looks like the nth derivative evaluated at the center divided by n factorial. Great. So what we have to do is we have to evaluate these derivatives at the center, which the center is 0. Well, when you look at the at the 0th derivative, if you plug in 0, you're going to get just a 1, because 1 to any power is just k. Sorry, 1 to any power will just be 1, even if it's the kth power. With the first derivative, if you plug in 0, you're going to get 1 to the k minus 1. You'll just be left with a k. So the first derivative is k. If you plug in the second derivative, 0 takes away the x. You get 1 to any power is just 1, right? So this is the beauty here of sinuring this thing at 0. It doesn't matter, it doesn't matter what the power of 1 plus x is, is that when you plug in 0, you always get 1 to some power, which is equal to 1. And so the only thing that survived was the k and the k minus 1, which you see right here. And so for the third derivative, the coefficients will stick around. The 1 plus x will disappear. So you get the third derivative is k times k minus 1 times k minus 2. And then finally, for the nth derivative, you're going to get k times k minus 1 times k minus 2 all the way down to k minus n plus 1. The 1 plus x will just disappear, and you get this thing right here. And so this gives us a formula for the nth derivative of f evaluated at 0. We also needed n factorial in the bottom. So we're going to divide this by n factorial. And so what we do is we see the following formula right here. The coefficient of this, of this power series will be k, k minus 1 times k minus 2, all the way down to k and minus, or k minus n plus 1 divided by n factorial, which you see right here. And as this, this, this, this, this, this is in front of us right here, we're going to abbreviate using the following notation. You're going to put k over n and there's parentheses. The parentheses are necessary to describe this notation. It's not a fraction. There's no bar that goes between them, so no bar. So this is what's referred to as the binomial coefficient. Now, the reason it gets its name here is for the following. What if we took like 1 plus x cube, right? What if k was a positive number? Well, you end up with 1 plus x times 1 plus x times 1 plus x. And what you can do is you can multiply this thing out. When you foil this thing, you would end up with 1 plus 3x plus 3x squared plus x cubed. And the coefficients of this sequence will look like the first number 1. This is actually 3, choose 0. The next one is 3, choose 1. The next one is 3, choose 2. And then the next last one is 3, choose 3. Where these numbers, these numbers k choose n, this actually looks like k times k minus 1. It's this formula right here. Sorry about that. Got a little messy. It's this formula right here. And because the thing is when your exponent k is a positive number, this will eventually go down to 0, right? This process eventually stops. It stops like right here. And you can actually compute the coefficients of this binomial expansion. And this is known as the binomial theorem from like an algebra class, right? So what we're doing is we're extending the binomial theorem to include powers beyond just positive integers, right? So instead of like 1 plus x cube, we're going to get a formula for like 1 plus x to the negative 3 or 1 plus x to the 1 half power. Our power of k doesn't have to be a positive integer any more. And so because of that connection to the binomial theorem from algebra, this series is commonly referred to as the binomial series. So f of x, which is equal to 1 plus x to the k, it'll be equal to its Maclaurin series, which is known as the binomial series. And you'll get the sum where n equals 0 to infinity of k over n of k choose n x to the n. This number right here is often called k choose n. k choose n. And the reason we had that before is that this is a this is a win. Again, when k is a positive integer here, you can use this to count things like, oh, I have four, you know, I have four donuts on the table and I choose to eat two of them. How many ways can I choose four donuts from two, right? So you have two donuts, you want to you have four donuts total, you're choosing four of them. How many ways can you do it? Well, you're going to get four, I guess I wrote backwards, I'm sorry, this would be, I was right the first time four choose two. So you get four choose two, and you can see there's going to be four times three over two times one, in which case you get six ways you can choose two donuts amongst the four to eat. And so that's where it gets us k choose four. This is a generalization of this binomial coefficient, this binomial theorem, and hence we call it the binomial series. This is a formula you're going to want to record later, so you can talk about these binomial series, which are pretty important. Now, with all the emphasis I put on Taylor's inequality, you might wonder, is the is the function actually equal to its binomial series? And the answer to this is that it's going to be yes. The function is equal to its binomial series. How do we see that? Well, one could do an argument, we have to determine what's the radius of convergence, first of all. One could do an argument with the ratio test a n plus one over a n, you can see that right here. If you take a n plus one, you get this thing right here. If you take one over a n, you get this thing right here. Putting the factorials together, you get n factorial over n plus one factorial, that'll simplify just to be one over n plus one. If you take this k, k minus one, k minus two, all the way to k minus n, and you divide that by k, k minus one, k minus two, all the way to k minus n plus one, most of those things will cancel out, right? You get k cancel, k minus one cancels, most of these things are going to cancel out with exception of, whoops, I canceled the wrong one, everything except for the k minus n should cancel out there. As soon as I did all those things correctly, no, I think I, I think I, I think I botched that. The one on, let's see, the one on, which is the bigger one, the one on top is the biggest one, right? Because k minus n plus one, that's the biggest number, so it didn't have anything to cancel out there, so there should be like a plus one in there. It looks like it's a typo. And so then the x's will also cancel out, leaving you just an x. So you see something like you see here before, although there's a coefficient off, that's okay. You get one over n plus one times k minus n plus one times x, and so you get a plus one right here again. Now taking the limit, this typo seems to exist forever for the rest of this thing. So as you take the limit as n goes to infinity, you're going to see that this thing is just equal to the absolute value of x, things will cancel out. And I can show you some more, again, there's some typos here, so don't worry about that too much, I'll fix these in the lecture notes that you can see linked to this video. But this thing will go off towards x, and this is with the ratio test, so we need x to be less than one. This tells us that the radius of convergence is one, so the function only is, well the radius of convergence for the series here is going to equal one. There are some exceptions to that, of course. What happens when x equals one? What happens when x equals negative one? It kind of depends on the k value. It depends on k. Of course, like if k is a positive integer, actually the radius of convergence is infinity because it's just a polynomial in that case, but for other values, it'll be one. And whether it converges at one or negative one depends on k. And we'll see an example of this just a moment. And you can use Taylor's inequality to show you that the function is equal to the binomial series. What I want to do here is do an example of actually computing the binomial series for a specific function. Let's take the function f of x equals one over the square root of four minus x. And I do want to convince ourselves that this is a binomial series here. So f of x equals one over the square root of four minus x. Notice that you can first of all factor out of four, right? Because with the binomial series you want a one plus x. We have a four minus x. How do you deal with that? So we're going to factor out of four that leaves behind one minus. We want a plus there. So we're going to get one plus a negative x over four. This all sits below one. The square root of four is two. So we get a one half that sits out in front. And then the square root is the one half power. So we're going to get one plus a negative x over four. And then the square root is the one half power. And since it's a denominator, we get a negative sign. So we're going to get a negative one half right here. And so this tells us we have a binomial series with a k value of negative one half. And we're going to substitute u with a negative x over four into our series. So our function f of x is going to equal one half times the sum where n ranges from zero to infinity. We're going to get the binomial coefficient negative one half choose n. And then we're going to get negative x over four raised to the nth power. If we plug that into the formula right there. And this right here gives us the series we could stop here if we really wanted to. But if you unravel the binomial coefficient a little bit, what you're going to see is the following. You're going to get one half. Then you're going to times that by, again if you write this, write this thing in expanded form, you're going to get one, right? One thing you should know about binomial coefficients is that if you take k choose zero, k choose zero, this will look like, well you, what was the original formula again, right? You're supposed to take no terms, you have no terms, you get a one over zero factorial, which is going to just equal one. k choose zero is always a one. k choose one, that's going to be k over one factorial, that's always just equal to k. Other ones are a little bit more difficult to do, but like if you do k choose two, you end up with k times k minus one over two factorial. I just kind of continue with that pattern there. So you're going to end up with one plus negative one half you get from the binomial coefficient. Then you're going to get a negative x over four. We'll come back to that one just a moment. Then the next term you're going to get k choose two, which that thing is going to look like negative one half on top. You're then going to get negative three halves for k minus one. This sits above two factorial, which is just two, times that by negative x over four squared. The next term we would get negative one half times negative three halves times negative five halves. This last term is k minus two. This sits above six, which is three factorial. And then we're going to get negative x over four cubed. And this pattern will continue on and on and on. I think we have enough to figure out what's going on here. So notice some things that are going to happen. Again, the one half sits out in front. We get this one. We get minus. We'll actually get plus negative negative, right? We get plus. We're going to get x on top and eight on bottom. The next one we're going to get minus, minus, and then and then so the negatives here cancel out. And then you're squaring a negative. So it's also a plus. So you're going to get a plus. You're going to get one times three, which is three. You're going to get an x squared. And on the bottom, you end up with a two. You get a two. You get a two. So that's eight. And then you get four squared, which is 16, right? So in that situation, what do we have? We get four times eight times 16. That looks like a big one. We'll come back to that one later. For the next term, what's going to happen here? You're going to get negative, negative, negative times negative, negative, negative, right? That's going to be a positive as well. You're going to end up with a one times three times five. So that's a 15 on top. You're going to get x cubed. And then you're going to get a two, a two, a two. So that's, there's eight there. You're going to get four cubed. That's a big number times six. So you get like eight times six times four cubed. And this kind of continues on and on and on. And so again, trying to describe the general solution, because you can also distribute this one half through if you want to, but I'm going to leave it out factored. So in general, if you want to describe this without the binomial coefficient, this would look something like n equals zero to infinity. What you're going to see at the top is you see this factorial like object, one times two times five times, keep on going. But what you do is you don't get every number like a factorial does. You actually end up with only odd numbers. So you get like this odd factorial, one times three times five times seven times nine times 11 up to wherever you terminate at, which will be two n minus 11. In the denominator, you're going to end up with an n factorial. That came from like the two factorial, the three factorial we see right there. But what else are we getting here? We're getting a bunch of twos that are popping up everywhere, because every time we do another k minus whatever, we're going to get a multiple of one half, which those gather together to make some eights. But we also can combine it together. We can also combine it together with the powers of four we have. And so in the end, it turns out that you're ending up with eight to the n in your denominator x to the n. So if you keep track of all the twos there, we should have eights or like what we had here is you had an eight, right? And keep on going from that. And so then our formula in the end should look something like this. Again, I went through that kind of a little bit quickly, because honestly, when it comes to these binomial coefficients, really, really, what we like to do is we just like to use the binomial coefficient formula, right? I mean, that's what we have. That's what we have this abbreviation to keep track of all these things. This is a pretty, this is a perfectly good way of expressing it. If you want to do it in expanded form, you can do something like this. I went through it kind of quickly. So I apologize if someone got a little bit confused. The intention isn't necessarily to get to this formula. And I'm actually quite good, quite, quite happy with everyone using this formula right here. This is perfectly fine. What we need to finish off, though, is we actually have to deal with, I mean, it's kind of interesting to see that this thing's actually not an alternating series, because you have this negative one to the n right here, you think that's alternating, but this negative k here also is alternating, so they actually balance each other out. What we want to finish off with is with the radius of convergence, because what we saw before is that negative x over four, much like a geometric series, it needs to be less than one. And so this would imply that the absolute value of x over four is less than one. So this tells us that the absolute value of x should be less than four, which this is our radius of convergence. So this binomial series will converge with the radius of convergence of four. And so that shows us how one can deal with a binomial series.