 In this video, we're gonna discuss the solution to question 15 from the practice final exam for math 12-20. And in this question, we're asked to compute, well to find the Taylor series for the function f of x equals the natural log of x centered at a equals one. So one thing to remember when it comes to Taylor series is that the formula for the Taylor series will be the sum where n ranges from zero to infinity. You're gonna get the nth derivative of the function evaluated at the center a divided by n factorial. And you can find times that by x minus a to the n. So we need to start calculating derivatives of this function. So if we take the zero derivative, that's just the function itself, which is the natural log of x. We divide that by, well not divide, we take the first derivative, we're gonna get one over x. We're gonna take the second derivative and so by the power rule, or if you're really into it right now, you could do the quotient rule, but by the power rule, because you're thinking this is x to the negative one right here, you're gonna get negative one over x squared. The next derivative, we have to calculate enough so that we can identify the pattern here. The next derivative, if you want, this would be negative x to the negative two. You're going to get positive two x to the negative three, or I'll write this as two over x cubed. And this would equal the same thing as two x and negative three. And you're gonna see this pattern continues on as you take the fourth derivative. You're gonna get two times three, which is, yeah, that's a four right there if you can't tell four. So the next one we're gonna get one times two times three over x to the fourth and then it's gonna be negative in the situation. If we do the fifth derivative, you're gonna end up with a positive one times two times three times four over x to the fifth. And so in general terms, notice what's happening here. You're gonna get for the nth derivative of x. It seems to be alternating. It goes positive, negative, positive, negative, positive. So it's offset by one. So when you have an odd derivative, it's gonna be a positive number. So that's gonna be a negative one to the n plus one or minus one if you prefer, something like that would work. And then in the numerator, you seem to have a factorial going on there. One, two, three, four. It's always one less than where we started. So the power on the bottom is gonna be x to the first derivative had a first power, second derivative had a second power, third derivative had a third power. So you're gonna get x to the end of the denominator and they have an n minus one factorial in the top. This gives us the general formula for the derivative so long as n is greater than or equal to one because the zero derivative is just the natural log of this. And so for the Taylor series, we need to just, for the Taylor series of the natural log of x, we're gonna get that this is equal to the first term, which is the natural log of one. And then we're gonna add to that the sum starting at a equals one. We're gonna separate the zero term from the rest because the zero term doesn't quite match our pattern but after we get to the first derivative does seem to match a very nice pattern. We're gonna end up with, if we plug in these formulas, right, the nth derivative at one divided by n factorial times x minus one to the n as we go off towards infinity. Now some things to mention here is that the natural log of one is zero. So that's just gonna disappear. If we evaluate this function, the nth derivative at one, you're gonna end up with negative one to the n plus one. You're gonna get this n minus one factorial but then you're gonna get this one to the n. One to the n is just, of course, of one. And so making some of these simplifications here, our Taylor series is gonna look like the sum where n equals one to infinity. We're going to get a negative one to the n plus one. We're gonna get this n minus one factorial on top. We have this n factorial on the bottom, x minus one here to the n. And then finally, the n factorials do cancel a little bit because n can factor as n times n minus one factorial for which those cancel. And so in the end, we get that the natural logs, it's Taylor series expansion at one is the sum where n equals one to infinity. We get this alternating sum. So we're gonna get negative one to the n plus one over n times x minus one to the nth. And so this provides a way where we can compute the Taylor series expansion for this natural log function at one. It doesn't have a McLaren series because we can't expand the natural log of zero because it's undefined at zero. We have a vertical asymptote at zero and we could expand it at one. And this works out here. This question is not asking you actually to find the radius of convergence for this thing. If you're sending it at one, the best you would expect would be the radius of convergence to be one itself because expanding to the left as far as you go would be at zero, right? You can't get past zero. And you actually can show that the radius of convergence here is one. You don't have to provide the interval convergence or the radius of convergence on this one. You also don't have to prove that the function's equal to its Taylor series. It's McLaren series. That is a fact that you can use here, but that's not what this question's asking. It's just asking to compute the Taylor series of a function. And this is by using the original formula we see right here. And this one that you'll see on the final, it'll be a function which has a predictable derivative pattern of some kind, like the natural log is a good example. Some other good ones to consider will be like cinch or cauch, the hyperbolic functions because the derivative of cinch is cauch so computing the McLaren series of cinch or cauch, for example, would be very, very similar to how we did that for sine and cosine. So those ones would be some other practice ones to try out on this one if you wanna see some other examples.