Differentiating power series | Series | AP Calculus BC | Khan Academy

So 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=0 ). And like always, pause this video and see if you can work it out on your own before we do it together.

All right, 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 we 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.

So let me do it the second way first. So let me just expand it out. ( f(x) ) is equal to... let's see, when ( n ) is equal to 0, this is ( (-1)^0 ), which is just one, times ( x^{0+3} ), so it's going to be ( x^{3/2} \cdot 0! ) which is just going to be over 1.

Then the next term, when ( n ) is equal to 1, well now it's going to be ( -1^1 ), so now we're going to have a negative out front. Negative, and it's going to be ( 2 \cdot 1 + 3 ), so that's going to be ( x^{5} \cdot 1! ) over ( 2 \cdot 1 + 1 ), so that's going to be ( 3! ), so it's going to be ( \frac{x^5}{6} ).

And then when ( n ) is equal to 2, this is going to be positive again, it's going to be ( x^{7} ) over ( 5! ). Is that right? Yeah, ( 5! ) and ( 5! ) actually, let me just write that out. ( 5! ) would be, what? ( 120 ). It'd be ( 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 ), so it would 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. So ( f'(x) ) is going to be equal to... we're just applying the power rule here. It's going to be ( 3x^{2} - \frac{5}{6}x^{4} + \frac{7}{5!}x^{6} ). 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... applying the power rule again, it's going to be ( 6x^{1} - \frac{20}{6}x^{3} + \frac{42}{5!}x^{5} ). And we're just going to keep on going, minus plus, keep going or alternate between minus something and then plus something on and on and on forever.

And then we get to the third derivative. The third derivative is equal to... let's see, derivative of ( 6x ) is ( 6 ) and then we have ( -\frac{60}{6}x^{2} + \frac{5 \cdot 42}{5!}x^{4} - \cdots ) over and over and over again.

Then we just evaluate this at zero. ( f'''(0) ): well, when ( x ) is equal to zero, all of these terms with ( x ) are going to go to zero, and you're just going to be left with this ( 6 ) here. So ( f'''(0) ) is just equal to ( 6 ).

Now another way that we could have tackled this is just kept it in this sigma notation. So we could have 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 have said ( f'(x) ) is equal to the sum from ( n = 0 ) to infinity, and you take the derivative here. You're going to 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 going to tell us which term we're... is going to tell us how we change from term to term.

So if I take the derivative with respect to ( x ) here, use the power rule: bring the ( 2n + 3 ) out front. So it's going to be ( (-1)^n \cdot (2n + 3) \cdot x^{2n + 2} ) over ( (2n + 1)! ).

And then if you want to 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 ( 0 ) to infinity of ( (-1)^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 going to have ( (2n + 3) \cdot (2n + 2) ) all of that over ( (2n + 1)! ) and that's going to be times ( x^{2n + 1} ).

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 ( 2n + 2 - 1 ) is ( 2n + 1 ).

And then if I want to take the third derivative, the third derivative is the sum from ( n = 0 ) to infinity of ( (-1)^n ). We take this, bring it, multiply it, so we're going to have ( (2n + 3) \cdot (2n + 2) \cdot (2n + 1) ) all of that over ( (2n + 1)! ), and then that is going to be times ( x^{2n} ).

Now let's now evaluate this thing when ( x ) is equal to zero. So ( f'''(0) ) is going to be the sum from ( n = 0 ) to infinity of ( (-1)^n ). Now this is going to be interesting: we're going to have all this business ( (2n + 3)(2n + 2)(2n + 1) ) all of that over ( (2n + 1)! ) times ( 0^{2n} ).

Now you might be tempted to say, well hey, if ( 0 ) to these different powers, maybe everything's going to be zero. But remember, we're starting at ( n = 0 ), so for any ( n ) that's not equal to zero, this ( 0^{2n} ) is just going to be zero and that term is going to disappear, kind of like what we saw when we expanded it out.

And so the only term that matters here is when ( n ) is equal to ( 0 ). And so this is just going to be equal to ( 0 ) for ( n = 1, 2, 3, 4, 5 ) all the way to infinity. This thing is going to dominate; it's just going to multiply; it's going to be zero and you're just going to zero everything out.

And so this is just going to reduce to the first term when ( n = 0 ). And when ( n = 0 ), it's going to be ( (-1)^0 ), which is just ( 1 ).

So I'll just let me just write that as ( 1 \times \frac{3 \cdot 2 \cdot 1}{1!} ), and then times ( 0^{0} ), which is equal to ( 1 ). So this is equal to ( 1 ) and so this is equal to ( 6 ).

So 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.

And this technique is useful because you'll see it a lot in math, where you might want to 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.