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- [Instructor] We're told here that f(x) is equal to this infinite series, and we need to figure out what is the third derivative of f, evaluated at x equals zero. And like always, pause this video and see if you can work it out on your own before we do it together. Alright, so there's two ways to approach this. One is we could just take the derivative of this expression while it's in sigma notation. The other way you do it is we could just expand out f(x) and take the derivative three times, and see if we get an answer that, I guess, makes sense. Let me do it the second way first. Let me just expand it out. F(x) is equal to, let's see, when n is equal to zero, this is negative one to the zero, which is just one, times x, times x to the zero plus three, so it's gonna be x to the third over two times zero, so that's zero plus one factorial, so that's just going to be over one. Then the next term, when n is equal to one, well now it's gonna be negative one to the one, so now we're just gonna have a negative out front. Negative, and it's gonna be two times one plus three, so that's gonna be x to the fifth power over two times one plus one, so it's gonna be two plus one is three factorial. So it's gonna be x to the fifth over six. And then when x is equal to two, this is going to be positive again, and it's gonna be x to the seventh power over five factorial. Is that right? Yeah. Five factorial, and five factorial... Actually, let me just write that out as five factorial. Five factorial would be, what, it would be 120. It'd be five times four times six, so it'd be 120. But we could just keep going minus plus, and it goes on and on and on forever. Well now let's just take the derivatives. F'(x) is going to be equal to, we're still applying the power rule here, it's going to be three x squared minus 5/6x to the fourth, plus seven over five factorial x to the sixth, I'm just applying the power rule, minus plus, we just keep going on and on and on forever. The second derivative, f''(x) is going to be equal to, apply the power rule again. It's going to be six x to the first minus four times five over six, I'll just write that as 20/6x to the third, plus six times seven, so it's 42 over five factorial x to the fifth, and we're just gonna keep on going, minus plus, keep going, or alternate between minus something then plus something, on and on and on forever. Then we get to the third derivative. The third derivative is equal to, let's see, the derivative of six x is six, and then we have minus 20 times three is 60/6, which of course is 10, x squared, plus five times 42 is what, 210 over five factorial times x to the fourth power, minus plus over and over and over again, and then we just evaluate this at zero. F'''(0), well, when x is equal to zero, all of these terms with xes are gonna go to zero, and you're just gonna be left with this six here. So f'', the third derivative evaluated at zero is just equal to six. Now another way that we could've tackled this is just kept it in this sigma notation. We could've said that f'(x) is equal to the infinite sum, and actually, let me line them up. So this is where we did f'(x) expanded out, but we could've said f'(x) is equal to the sum from n equals zero to infinity, and you take the derivative here, you're gonna get, and you're taking the derivative with respect to x, so for that purpose, you assume everything else is, well, the n is just gonna tell us, is gonna tell us how we change from term to term, so if we take the derivative with respect to x here, use the power rule, bring the two n plus three out front, so it's gonna be negative one to the n times two n plus three, times x to the decrement, the exponent, two n plus two over two n plus one factorial. And then if you wanna take the second derivative, and this is the same thing as this. If you take the second derivative, f''(x), well now we're taking the sum from zero to infinity of negative one to the n. Let me move over to the right a little bit so we have some space. And now, we take this exponent out front, so you're gonna have two n plus three times two n plus two, all of that's going to be over two n plus one factorial, and this is gonna be times x to the two n plus one. All I'm doing every time, it seems really complicated, I'm just taking the exponent out front, multiplying it out front, and then decrementing it. So two n plus two minus one is two n plus one. And then if I wanna take the third derivative, the third derivative is the sum where n equals zero to infinity, negative one to the n. We take this, bring it, multiply it, so we're gonna have two n plus three times two n plus two, times two n plus one, all of that over two n plus one factorial, and then that is going to be times x to the two n power. Now let's now evaluate this thing when x is equal to zero. F'''(0) is gonna be the sum from n equals zero to infinity of negative one to the nth power. This is gonna be interesting. We're gonna have all of this business, two n plus three times two n plus two times two n plus one, all of that over two n plus one factorial, times zero to the two n power. You might be tempted to say, well, hey, if zero to these different powers, maybe everything's gonna be zero, but remember, we're starting at n equals zero, so for any n that's not equal to zero, this zero to that power is just gonna be zero and that term's gonna disappear, kinda like what we saw when we expanded it out. And so the only term that matters here is when n is equal to zero. So this is just going to be equal to, because for n equals one, two, three, four, five, all the way to infinity, this thing is gonna dominate. It's just gonna multiply, it's gonna be zero. You're just gonna zero everything out. And so this is just gonna reduce to the first term, when n equals zero, and when n equals zero, it's gonna be negative one to the zero. This is gonna be, which is just one. Let me just write that as one. Times, this is three times two times one over one factorial, and then times zero to the zero, which is equal to one. So this is equal to one, and so this is equal to six. Either way, I think the first way we did it was a little bit more straightforward, a little bit more intuitive, closer to what you might have seen before, but it's important to realize that we did the same thing both times, we just kept it in the sigma notation on this time to the right. This technique is useful because you'll see it a lot in math, where you might wanna do things a little bit more of a general way, and so it might be helpful to take the derivatives while you stay in that sigma notation.